acr ea unit 2 - apple

®

AccuplacerCollege-ReAdy Version 2.0

ElEmEntary algEbra UnIt 2

The classroom teacher may reproduce materials in this book for classroom use only.The reproduction of any part for an entire school or school system is strictly prohibited.

No part of this publication may be transmitted, stored, or recorded in any formwithout written permission from the publisher.

1 2 3 4 5 6 7 8 9 10

ISBN 978-0-8251-6773-7

Copyright © 2011

J. Weston Walch, Publisher

Portland, ME 04103

www.walch.com

Printed in the United States of America

EDUCATIONWALCH

Accuplacer College-Ready Mathematics: Elementary Algebra© 2011 Walch Educationiii

Unit 2 • operations with algebraic expressions

Table of Contents

iii

Teacher’s Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TG1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TG1

Station Activities Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TG3

Unit 2 Pre-Assessment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Lesson 1: Evaluation of Simple Formulas and Expressions . . . . . . . . . . . . . . . . . . . . . . 5

Lesson 2: Adding and Subtracting Monomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .30

Lesson 3: Adding Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .52

Lesson 4: Subtracting Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .74

Lesson 5: Multiplying Monomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .96

Lesson 6: Dividing Monomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .119

Lesson 7: Multiplying Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .142

Lesson 8: Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .165

Lesson 9: Simplifying Algebraic Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .188

Lesson 10: Dividing Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .210

Lesson 11: Radical Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .233

Unit 2 Mixed Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .261

Unit 2 Post-Assessment 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .265

Unit 2 Post-Assessment 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .269

Station Activities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .273

Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .319

067737E

Accuplacer College-Ready Mathematics: Elementary Algebra© 2011 Walch Educationtg1


Teacher’s GuideIntroductionThe Accuplacer College-Ready (ACR) Instructional Materials are a complete set of resources developed to support students, and their teachers, in preparation for the Accuplacer® test used by many colleges and universities to inform placement decisions. This program was developed around the content and format of the test, with input from educators familiar with mathematics and with the Accuplacer®. The ACR Materials comprise a full course, providing more than enough activities and resources for a semester of instruction that develops and reinforces the mathematics needed for the Accuplacer® and for success in future math courses. This work was funded in part by a Davis Family Foundation Grant, awarded to the Maine International Center for Digital Learning. The materials were piloted by teachers in MELMAC Learning Foundations College-Ready grant sites. They were revised and refined based on pilot feedback and are presented here as Version 2.0.

The ACR program recognizes the importance of tailoring instructional experiences to address students’ identified needs. The materials are designed in a mix-and-match model, allowing teachers to select appropriate lessons, activities, and components.

Each unit in the Accuplacer College-Ready materials contains the following elements:

• Unit Pre-Assessment • For each lesson in the unit:

• Lesson Pre-Assessment• Essential Questions• Words to Know• Warm-Up Option 1 with Debrief• Warm-Up Option 2 with Debrief• Focus Problem with Debrief• Additional Examples• Guided Practice• Independent Practice • Progress Assessment• Resource List• Answer Key

• Unit Mixed Review• Unit Post-Assessment 1• Unit Post-Assessment 2• Station Activities• Unit Glossary

Unit 2 • operations with algebraic expressionsTeacher’s Guide

Accuplacer College-Ready Mathematics: Elementary Algebra tg2

© 2011 Walch Education

All of the program materials are completely reproducible and may be printed or projected to support instruction. The complete set of materials encompasses six units, focused on six of the Accuplacer® content strands. Each unit provides overview materials as well as a series of lessons.

This unit includes the following lessons:

Elementary Algebra Unit 2: Operations with Algebraic Expressions

Lesson 1: Evaluation of Simple Formulas and Expressions

Lesson 2: Adding and Subtracting Monomials

Lesson 3: Adding Polynomials

Lesson 4: Subtracting Polynomials

Lesson 5: Multiplying Monomials

Lesson 6: Dividing Monomials

Lesson 7: Multiplying Polynomials

Lesson 8: Factoring

Lesson 9: Simplifying Algebraic Fractions

Lesson 10: Dividing Polynomials

Lesson 11: Radical Expressions

Structure of the Units

Unit and Lesson Pre-Assessments provide teachers with information to guide their instructional decisions and against which to document progress. Essential Questions serve to focus teaching and learning on important concepts. Words to Know (and the Unit Glossary) allow for direct vocabulary instruction, shown to increase mathematics learning and achievement. Two Warm-Up Options (with Debriefs) offer teachers choices for engaging students’ prior knowledge and for differentiating instruction. A Focus Problem (with Debrief) and Additional Examples facilitate problem-based learning, letting teachers present concepts and skills in meaningful, real-world contexts. Guided Practice and Independent Practice problem sets give students opportunities to hone their skills, and Progress Assessments gauge their progress after each lesson. The Resource Lists include links to online resources, with synopses to help teachers make decisions about their use. (Note: Functionality of some online resources may be improved by switching to another browser, such as Safari if using a Mac or Firefox if using a PC.) A Mixed Review and two Unit Assessments with Answer Keys complete each unit. Finally, Station Activities engage small groups of students in a series of problem-solving activities, with suggested debrief prompts.



Station Activities GuideEach unit includes a collection of station-based activities to provide students with opportunities to practice and apply the mathematical skills and concepts they are learning. You may use these activities in addition to the instructional lessons, or, especially if the pre-test or other formative assessment results suggest it, instead of direct instruction in areas where students have the basic concepts but need practice. The debriefing discussions after each set of activities provide an important opportunity to help students reflect on their experiences and synthesize their thinking. It also provides an additional opportunity for ongoing, informal assessment to guide instructional planning.

Implementation Guide

The following guidelines will help you prepare for and use the activity sets in this section.

Setting Up the Stations

Each activity set consists of four stations. Set up each station at a desk, or at several desks pushed together, with enough chairs for a small group of students. Place a card with the number of the station on the desk. Each station should also contain the materials specified in the teacher’s notes, and a stack of student activity sheets (one copy per student). Place the required materials (as listed) at each station.

When a group of students arrives at a station, each student should take one of the activity sheets to record the group’s work. Although students should work together to develop one set of answers for the entire group, each student should record the answers on his or her own activity sheet. This helps keep students engaged in the activity and gives each student a record of the activity for future reference.

Forming Groups of Students

All activity sets consist of four stations. You might divide the class into four groups by having students count off from 1 to 4. If you have a large class and want to have students working in small groups, you might set up two identical sets of stations, labeled A and B. In this way, the class can be divided into eight groups, with each group of students rotating through the “A” stations or “B” stations.


Accuplacer College-Ready Mathematics: Elementary Algebra tg4


Assigning Roles to Students

Students often work most productively in groups when each student has an assigned role. You may want to assign roles to students when they are assigned to groups and change the roles occasionally. Some possible roles are as follows:

• Reader—reads the steps of the activity aloud

• Facilitator—makes sure that each student in the group has a chance to speak and pose questions; also makes sure that each student agrees on each answer before it is written down

• Materials Manager—handles the materials at the station and makes sure the materials are put back in place at the end of the activity

• Timekeeper—tracks the group’s progress to ensure that the activity is completed in the allotted time

• Spokesperson—speaks for the group during the debriefing session after the activities

Timing the Activities

The activities in this section are designed to take approximately 10 minutes per station. Therefore, you might plan on having groups change stations every 10 minutes, with a two-minute interval for moving from one station to the next. It is helpful to give students a “5-minute warning” before it is time to change stations.

Since each activity set consists of four stations, the above time frame means that it will take about 50 minutes for groups to work through all stations.

Guidelines for Students

Before starting the first activity set, you may want to review the following “ground rules” with students. You might also post the rules in the classroom.

• All students in a group should agree on each answer before it is written down. If there is a disagreement within the group, discuss it with one another.

• You can ask your teacher a question only if everyone in the group has the same question.

• If you finish early, work together to write problems of your own that are similar to the ones on the activity sheet.

• Leave the station exactly as you found it. All materials should be in the same place and in the same condition as when you arrived.



Debriefing the Activities

After each group has rotated through every station, bring students together for a brief class discussion. At this time, you might have the groups’ spokespersons pose any questions they had about the activities. Before responding, ask if students in other groups encountered the same difficulty or if they have a response to the question. The class discussion is also a good time to reinforce the essential ideas of the activities. The questions that are provided in the teacher’s notes for each activity set can serve as a guide to initiating this type of discussion.

You may want to collect the student activity sheets before beginning the class discussion. However, it can be beneficial to collect the sheets afterward so that students can refer to them during the discussion. This also gives students a chance to revisit and refine their work based on the debriefing session. If you run out of time to hold class discussions, you might want to have students journal about their experiences and follow up with a class discussion the next day.

Unit 2 • operations with algebraic expressionsUnit Pre-Assessment

naMe:

Accuplacer College-Ready Mathematics: Elementary Algebra© 2011 Walch Education1

Assessment

Simplify the following expressions. Circle the letter of the best answer. Show your work.

1. −45ab

, where a = –10 and b = 6

a. –120 c. 48

b. –48 d. 120

2. 6a2 – 3b2 – 2ab + 10, where a = –1 and b = 2

a. –12 c. 0

b. –4 d. 8

3. 26x + 12 – 13 – 2x + 6x – 3

a. 18x – 4 c. 30x – 4

b. 27x – 1 d. 32x – 6

4. (x + 5) + (6 – 3x) + (–10x + 3)

a. –12x + 8 c. –6x + 14

b. –6x + 2 d. –12x + 14

5. (3x – 5x2 + 6) – (x2 + 9x – 14)

a. –6x2 – 6x + 20 c. –4x2 – 6x + 20

b. –6x2 + 6x – 8 d. –4x2 + 12x – 8

6. a3b3c5d5 • ab2c3 • c3d2 • d5c2a

a. a4b5c13d12 c. a5b5c13d12

b. a5b5c13d14 d. a5b5c16d12

continued


naMe:

Accuplacer College-Ready Mathematics: Elementary Algebra 2


Assessment

7. m3n–5p2q–1 ÷ p–1q–5m3n6

a. n–11pq–6 c. npq–4

b. n–11p3q4 d. m6n–5p2q–1

8. (x2 + y)(x2 + 2xy + y2)

a. x3 + 3x2y + 3xy2 + y3 c. x4 + 2x3y + 2x2y2 + 2xy3 + y4

b. x4 + x2y2 + 3x2y + 2xy2 + y3 d. x4 + 2x3y + x2y2 + x2y + 2xy2 + y3

9. (y + 4)(y – 4)(y2 + 16)

a. y4 – 16 c. y4 + 32y2 + 256

b. y4 – 256 d. y4 – 8y3 + 32y2 – 128y + 256

10. Factor 16x2 – 225.

a. (4x + 15)2 c. (4x – 15)2

b. (4 x – 15)(4x + 15) d. This expression cannot be factored.

11. 15 24 9

3

3 2x x xx

+ −−

a. –5x2 – 8x – 3 c. –5x2 – 8x + 3

b. –5x2 + 8x + 3 d. –5x2 + 8x + 3

12. (8x3 + 48x2 + 96x + 64) ÷ (2x + 4)

a. 4x2 + 16x + 16 c. 4 16 161

2 42x x

x+ + +

+

b. 4 1616

2 42x x

x+ +

+ d. 4 16 16

642 4

2x xx

+ + ++

continued


naMe:


Assessment

13. (5x4 + 3x2 + 6x – 20) ÷ (x – 1)

a. 5 8 1461

2x xx

+ + −−

c. 5 5 8 1461

3 2x x xx

+ + + +−

b. 5 5 8 1461

3 2x x xx

+ + + −−

d. 5 10 13 1911

3 2x x xx

+ + + −−

14. x x x+( ) −( ) − +3 3 92

a. 0 c. x2 – x

b. x2 + 6 d. x2 3 3+ −

15. y

y

16

a. 4 c. 4 y

b. 4y d. 4 y

y




Unit Pre-Assessment Answer Key 1. c (Lesson 1) 9. b (Lesson 7) 2. d (Lesson 1) 10. b (Lesson 8) 3. c (Lesson 2) 11. c (Lesson 9) 4. d (Lesson 3) 12. a (Lesson 10) 5. a (Lesson 4) 13. b (Lesson 10) 6. c (Lesson 5) 14. c (Lesson 11) 7. b (Lesson 6) 15. c (Lesson 11) 8. d (Lesson 7)

Unit 2 • operations with algebraic expressionsLesson 1: Evaluation of Simple Formulas and Expressions

naMe:


Assessment

Lesson Pre-AssessmentEvaluate each expression if x = 3, y = –10, and z = 8. Show your work.

1. –6y + 4x2 – 3z + (10 – 3)

2. 52

2

2

xy

Write and evaluate an expression for each of the following problems.

3. A cookie-of-the-month club costs $12 per month for 1 dozen cookies. Additional cookies purchased that month cost $9.75 per dozen. How much will a customer in the club have spent after 1 month if he or she purchases 2 dozen extra cookies?

4. The length of a rectangle is twice its width. If the length is 10 feet, what is the area of the rectangle?

5. A circle has a radius of 3 inches. What is the circumference, in terms of π, of a circle with twice the radius of the original circle?




Instruction

Lesson 1: Evaluation of Simple Formulas and Expressions

Essential Questions 1. How are variables used to represent numbers?

2. How are algebraic expressions different from algebraic equations?

3. How is the order of operations applied to expressions and simple formulas at specific values?

WORDS TO KNOW

coefficient the number multiplied by a variable in an algebraic expression

constant a quantity that does not change

expression a symbol or combination of symbols representing a value or relation

order of operations the order in which expressions are evaluated from left to right (parentheses, exponents, division and multiplication, and addition and subtraction—PEMDAS)

variable a letter used to represent a value that can change or vary


naMe:


Warm-Up Option 1Evaluate each expression.

1. 4 – 2 + 3

2. 10 ÷ 5 • 3 + 2

3. 10 – 7 + 32 – (4 + 16)

Write the expression that describes each of the situations below. Then determine which operation should be applied first to evaluate the expression. Lastly, evaluate each expression.

4. The length of a wooden board is 2 feet. A second wooden board has a length of 3 feet less than 20 feet. What is the total length of the boards if they are aligned end to end?

5. The perimeter of a rectangle is 20 feet. The perimeter of a second rectangle is 4 feet less than twice the perimeter of the first rectangle. What is the perimeter of the second rectangle?


naMe:



Warm-Up Option 2A hot dog vendor at a county fair sold 102 hot dogs on Monday. He sold twice that amount on Tuesday. On Wednesday, it rained and he sold 54 fewer hot dogs than he did on Monday. On Thursday, he sold 150 hot dogs, and on Friday, he sold half as much as he sold Thursday.

1. Write an expression that will calculate the total number of hot dogs he sold Monday through Friday.

2. Evaluate the expression from problem 1 to find the total number of hot dogs he sold Monday through Friday.



Instruction

Warm-Up Option 1: Debrief 1. 4 – 2 + 3

• Use the order of operations, PEMDAS, from left to right to evaluate the expression.

• The first operation is subtraction; find the difference of 4 and 2.

4 – 2 + 3

2 + 3

• The next operation is addition; find the sum of 2 and 3.

2 + 3

5

4 – 2 + 3 = 5

Many students have the misconception that addition always occurs before subtraction. Remind students that addition and subtraction are considered to have the same priority level and should be completed left to right.

2. 10 ÷ 5 • 3 + 2


• The first operation is division; find the quotient of 10 and 5.

10 ÷ 5 • 3 + 2

2 • 3 + 2

• The next operation is multiplication; find the product of 2 and 3.

2 • 3 + 2

6 + 2

• Finally, add 6 and 2.

6 + 2

8

10 ÷ 5 • 3 + 2 = 8




Instruction

Many students have the misconception that multiplication always occurs before division. Remind students that multiplication and division are considered to have the same priority level and should be completed left to right.

3. 10 – 7 + 32 – (4 + 16)


• Evaluate the parentheses first.

10 – 7 + 32 – (4 + 16)

10 – 7 + 32 – 20

• Evaluate the exponent.

10 – 7 + 32 – 20

10 – 7 + 9 – 20

• Add and subtract from left to right.

10 – 7 + 9 – 20

3 + 9 – 20

12 – 20

–8

10 – 7 + 32 – (4 + 16) = –8



Instruction

4. The length of a wooden board is 2 feet. A second wooden board has a length of 3 feet less than 20 feet. What is the total length of the boards if they are aligned end to end?

• Write an expression for each board, then sum the expressions to find the total length.

• The expression for the first wooden board is 2 ft.

• The expression for the second wooden board is (20 ft – 3 ft).

• The expression for the total length of the two boards is 2 ft + (20 ft – 3 ft).

• Use the order of operations, PEMDAS, from left to right on the expression.

• The first operation is to evaluate the parentheses: (20 ft – 3 ft) = 17 ft.

2 ft + 17 ft = 19 ft

The total length of the boards is 19 feet.

5. The perimeter of a rectangle is 20 feet. The perimeter of a second rectangle is 4 feet less than twice the perimeter of the first rectangle. What is the perimeter of the second rectangle?

• Write an expression for the perimeter of the first rectangle.

• The expression for the perimeter of the first rectangle is 20 ft.

• Use this expression and the given information to determine the perimeter of the second rectangle.

• Write an expression for the second rectangle.

• The expression for the perimeter of the second rectangle is (2 • 20 ft) – 4 ft.


• The first operation is to evaluate the parentheses: (2 • 20 ft) = 40 ft.

(2 • 20 ft) – 4 ft

= 40 ft – 4 ft

= 36 ft

The perimeter of the second rectangle is 36 feet.




Instruction

Warm-Up Option 2: DebriefA hot dog vendor at a county fair sold 102 hot dogs on Monday. He sold twice that amount on Tuesday. On Wednesday, it rained and he sold 54 fewer hot dogs than he did on Monday. On Thursday, he sold 150 hot dogs, and on Friday, he sold half as much as he sold Thursday.

1. Write an expression that will calculate the total number of hot dogs he sold Monday through Friday.

• Write an expression for the number of hot dogs sold each day and then find their sum.

• Monday = 102 hot dogs

• Tuesday = 2(102) hot dogs

• Wednesday = 102 – 54 hot dogs

• Thursday = 150 hot dogs

• Friday = 150 ÷ 2 hot dogs

The expression representing the total hot dogs sold for the week is 102 + 2(102) + 102 – 54 + 150 + 150 ÷ 2.

2. Evaluate the expression from problem 1 to find the total number of hot dogs he sold Monday through Friday.


102 + 2(102) + 102 – 54 + 150 + 150 ÷ 2

The parentheses here are used to indicate multiplication.

Exponents are not used in this problem.

• Multiplication: 2(102) = 204

102 + 204 + 102 – 54 + 150 + 150 ÷ 2

• Division: 150 ÷ 2 = 75

102 + 204 + 102 – 54 + 150 + 75

• Addition and subtraction: 102 + 204 + 102 – 54 + 150 + 75 = 579

579 hot dogs were sold.


naMe:


Focus ProblemThe revenue for tennis racquets sold at a sporting goods store each month is based on the sales and price of three different brands of racquets. The revenue for the tennis racquets is $150 for each Prince® racquet sold, plus $175 for each Wilson® racquet sold, plus $205 for each Head® racquet sold.

1. In this situation, what quantities are variables?

2. What expression represents the revenue for the number of Prince® racquets sold each month using the variable p?

3. What expression represents the revenue for the number of Wilson® racquets sold each month using the variable w?

4. What expression represents the total revenue for the number of Head® racquets sold each month using the variable h?

5. What expression represents the total revenue for the number of Prince®, Wilson®, and Head® racquets sold each month if twice as many Wilson® racquets were sold than Prince® racquets, and 30 fewer Head® racquets were sold than Prince® racquets?

6. Using the expression found in problem 5, find the total revenue if 50 Prince® racquets were sold.




Instruction

Focus Problem DebriefIntroduction

• To introduce variables, ask students how variables relate to numbers.

• Explain that variables are used to represent numbers.

• Use the following example to help explain this concept: The amount of rainfall each day of the month is variable. However, the actual amount of rainfall can be recorded each day.

• To introduce the difference between algebraic expressions and algebraic equations, explain that algebraic expressions do not have an equal sign.

• Write the following example on the board:

Example: 2x is an expression.

2x = 10 is an equation.

For the expression 2x, x can be any value.

For the equation 2x = 10, x can only be the value 5 since only 2(5) = 10.

• To introduce how to apply the order of operations for simple formulas and expressions at specific values, explain that the specific values for the variables must be substituted or “plugged in” to the formula or expression first. Then use the order of operations to evaluate the numeric formula or expression.

Write the following examples on the board:

Example 1

Evaluate 2x + (3y – 2) – 4(2z) + 10, where x = 1, y = 4, and z = 3.

Solution

• Substitute the specific values for the variables.

2(1) + (3(4) – 2) – 4(2(3)) + 10

• Be sure students recognize that 2x is equal to 2 times x.

• The substitution of 1 for x results in 2 • 1, not 21.



Instruction


• Begin with parentheses. Students often have difficulty matching parentheses. Identify a set of parentheses by determining the open and close parentheses. It is often helpful to start within a set of parentheses and match them up.

2(1) + (3(4) – 2) – 4(2(3)) + 10

2(1) + (12 – 2) – 4(2(3)) + 10

2(1) + 10 – 4(2(3)) + 10

2(1) + 10 – 4(6) + 10

• Evaluate all multiplication and division from left to right.

2(1) + 10 – 4(6) + 10

2 + 10 – 24 + 10

• Evaluate all addition and subtraction from left to right.

2 + 10 – 24 + 10

12 – 24 + 10

–12 + 10

–2

The expression is equal to –2.

Example 2

The formula for the perimeter of a rectangle is P = 2l + 2w. What is the perimeter of a rectangle with a length of 5 inches and a width of 3 inches?

Solution

• Identify known values: length = l = 5; width = w = 3

• Substitute the values for the variables.

P = 2(5) + 2(3)




Instruction


P = 10 + 6 = 16

The perimeter of the rectangle is 16 inches.

Focus Problem statement:

The revenue for tennis racquets sold at a sporting goods store each month is based on the sales and price of three different brands of racquets. The revenue for the tennis racquets is $150 for each Prince® racquet sold, plus $175 for each Wilson® racquet sold, plus $205 for each Head® racquet sold.

Question 1

In this situation, what quantities are variables?

Instruction

• Determine which quantities vary from month to month.

• The price of each racquet is constant, or remains the same, each month.

• The quantities that vary each month are the number of Prince®, Wilson®, and Head® racquets sold.

Question 2

What expression represents the revenue for the number of Prince® racquets sold each month using the variable p?

Instruction

The revenue is the price of the Prince® racquet multiplied by the number of Prince® racquets sold.

This is represented by the expression 150p.



InstructionQuestion 3

What expression represents the revenue for the number of Wilson® racquets sold each month using the variable w?

Instruction

The revenue is the price of the Wilson® racquet multiplied by the number of Wilson® racquets sold.

This is represented by the expression 175w.

Question 4

What expression represents the total revenue for the number of Head® racquets sold each month using the variable h?

Instruction

The revenue is the price of the Head® racquet multiplied by the number of Head® racquets sold.

This is represented by the expression 205h.

Question 5

What expression represents the total revenue for the number of Prince®, Wilson®, and Head® racquets sold each month if twice as many Wilson® racquets were sold than Prince® racquets, and 30 fewer Head® racquets were sold than Prince® racquets?

Instruction

• Write the revenue for each brand of racquet as a separate expression.

Prince®: 150p

Wilson®: 175w

Head®: 205h

Total revenue: 150p + 175w + 205h




Instruction

• Identify the number of racquets sold.

Number of Prince® racquets sold (p): p

Number of Wilson® racquets sold (w): twice Prince® or 2p

Number of Head® racquets sold (h): 30 fewer than Prince® or p – 30

• Substitute the number of racquets sold into the expression for total revenue.

Total revenue: 150p + 175w + 205h

Total revenue: 150p + 175(2p) + 205(p – 30)

Question 6

Using the expression found in problem 5, find the total revenue if 50 Prince® racquets were sold.

Instruction

Number of Prince® racquets sold: 50, so p = 50.

• Evaluate the expression 150p + 175(2p) + 205(p – 30) for p = 50.

• Substitute 50 for p in the expression 150p + 175(2p) + 205(p – 30).

• It is helpful to place parentheses around substituted values, especially as problems become more complex.

150(50) + 175(2(50)) + 205((50) – 30)

• Use the order of operations to evaluate the expression.

• Evaluate the innermost sets of parentheses first.

((50) – 30) = (20)

(2(50)) = (100)

• Substitute these values into the expression.

150(50) + 175(2(50)) + 205((50) – 30)

150(50) + 175(100) + 205(20)



Instruction

• Perform multiplication from left to right.

150(50) = 7,500

175(100) = 17,500

205(20) = 4,100

• Substitute these values into the expression.

150(50) + 175(100) + 205(20)

7,500 + 17,500 + 4,100

• Perform addition from left to right.

7,500 + 17,500 + 4,100 = 29,100

The total revenue if 50 Prince® racquets were sold is $29,100.

Additional ExamplesExample 1

A cell phone plan includes a flat fee of $40 per month plus $0.10 per minute used. Write an expression to represent the monthly cost of the cell phone plan.

Solution

• Determine the variable amount and the constant.

Variable: number of minutes used

Constant: flat fee of $40 per month

• Let x = the number of minutes used.

Monthly cost = flat fee + the variable fee

Monthly cost = 40 + 0.10x

Expression for the phone’s monthly cost: 40 + 0.10x




InstructionExample 2

Based on the expression for Example 1, what is the cost of a cell phone plan that used 200 minutes this month?

Solution

• Use the expression 40 + 0.10x, with x = 200 minutes.

• Substitute 200 for x.

40 + 0.10(200)

• Evaluate the expression.

40 + 0.10(200)

= 40 + 20

The cost of the plan is $60.

Example 3

The formula for the surface area of a cylinder is A = 2πr2 + 2πrh, where r = radius and h = height of the cylinder. What is the surface area, in terms of π, of a cylinder with a radius of 4 inches and a height of 9 inches?

Solution

• Substitute the given values of r and h into the formula.

A = 2π(42) + 2π(4)(9)

• Use the order of operations to find the area of the cylinder.

A = 2π(42) + 2π(4)(9)

A = 2π(16) + 2π(4)(9)

A = 32π + 72π

A = 104π

The surface area of the cylinder is 104π square inches.


naMe:


Guided PracticeDetermine if each problem represents an expression or an equation.

1. 12x2 – 2y + 10(4)

2. 4y + 3x2(10 – 4) = 26

3. 12z ÷ 2 + 48y – 14

4. 5x + 2y = 45

Evaluate each expression if x = 2, y = –4, and z = 7.

5. xy

2

6. y(x – 4) + z(12 – 9)

7. xyz – 14

8. 4(10 – 6) + 3x – 2(z2 – 5)

continued


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Evaluate each simple formula given the values of the variables.

9. Evaluate A = πr2, in terms of π, where r = 8 inches.

10. Evaluate V = lwh, where l = 10 feet, w = 3 feet, and h = 4 feet.

Use the order of operations to evaluate the following expression.

11. The number of books sold at a bookstore over the past 5 days follows the expression 45 + 2(10) + 51 + 4(32) + 20(52 – 20). How many books were sold at the bookstore over the past 5 days? Show your work.

For problems 12–14, write an expression or simple formula that represents the given scenario, and then evaluate it.

12. A gym membership at Strong Arms Gym costs $45 per month. There is an initial fee of $30 to join the gym. How much money would a new member have spent on a gym membership after 6 months?

13. The area of a rectangle follows the formula A = lw. The length is 12 feet and the width is 23

the length. What is the area of the rectangle?

continued


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14. Cell phone company A charges $0.20 per minute with no flat fee per month. Cell phone company B charges $0.05 per minute with a flat fee of $30 per month. Write an expression and determine the cost of each company’s cell phone plan if 100 minutes are used.

For problems 15–16, use the order of operations to evaluate each expression.

15. The cost of gas per gallon on Monday was $2.50. On Tuesday it increased by $0.10 per gallon. On Wednesday it cost half the amount of the sum of the cost per gallon of Monday and Tuesday. If 5 gallons of gas were purchased each of the three days, what was the total cost of gas Monday through Wednesday?

16. A large circle has a radius of 10 inches. A circle within that circle, with the same center, has a radius of 4 inches. What is the area, in terms of π, of the shaded region?


naMe:



Independent Practice Determine if each problem represents an expression or an equation.

1. 3x2 – 2 = 25

2. 2y + 2x2 – 14 = 101

3. 4x ÷ 1 + 24 – (10 + x)

4. 2(9) + (15 + 42)

Evaluate each expression if x = 3, y = 1, and z = 10.

5. 2 42x

y+

6. y(10 – z) + 4(z + 5)

7. x + y + z – 26y

8. 8(33 – 9) + 2x – 4(z2 – 10)

continued


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Evaluate each simple formula given the values of the variables.

9. Evaluate A = πr2, in terms of π, where r = 4 inches.

10. Evaluate P = 2w + 2l where l = 20 feet and w = 8 feet.

Use the order of operations to evaluate the following expression.

11. The number of tickets sold at a movie theater over the past 5 days follows the expression 100 + 3(60) + 75 + 202 + 15(62 – 18). How many tickets were sold at the movie theater over the past 5 days? Show your work.

For problems 12–14, write and evaluate each expression or simple formula that represents the given scenario.

12. A dinner club membership at a local restaurant costs $30 per month. With the membership, fixed-price dinners cost $25 each. How much money would a member have spent on a dinner club membership after 4 months if he eats 5 fixed-price dinners per month?

13. The formula for finding the area of a triangle is A bh=12

. The base is 10 feet and the height is 14

base. What is the area of the triangle?

continued


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14. Anna’s Book Club charges $5.99 per book with no flat fee per month. Brad’s Book Club charges $9.99 per book with a flat fee of $5 per month. Write an expression and determine the cost of each book club if 10 books are purchased in 1 month.

For problems 15–16, use the order of operations to evaluate each expression.

15. The cost of milk per gallon on Monday was $3.27. On Tuesday it went on sale and decreased by $0.50 per gallon. On Wednesday it cost $1.00 more per gallon than what it cost on Tuesday. If 2 gallons of milk were purchased each of the three days, what is the total cost of milk Monday through Wednesday?

16. A large square has a side length of 12 inches. A square within that square, with the same center, has a side length of 9 inches. What is the area of the shaded region?


naMe:


Assessment

Progress AssessmentEvaluate each expression if a = 4, b = –2, and c = 9. Show your work.

1. 10c + 2a2 – 4ac

2. 244ab

Write and evaluate an expression for each of the following problems.

3. A coffee-of-the-month club charges $15 per month for 1 bag of coffee. Additional coffee purchased that month costs $12 per bag. How much will a customer in the club have spent after 2 months if she purchases 3 extra bags of coffee each month?

4. A tutoring service offers tutoring to non-club members at $75 per hour. The tutoring service offers tutoring to its club members at $25 per hour, but members must also pay a $20 per month flat fee. If a club member and a non-club member each have 5 hour-long tutoring sessions in 1 month, what will each person pay?

5. A circular campaign button has a radius of 2 inches. What is the area, in terms of π, of the button?

6. The perimeter of a triangle is found using the formula P = a + b + c, where a = 3 feet, b = 4 feet, and c = 5 feet. What is the perimeter of a triangle that is 3 times the size of the original triangle?




Instruction

Resource List• Alberta Education. “Order of Operations.”

www.learnalberta.ca/content/mec/flash/index.html?url=Data/1/A/A1A2.swf

This math interactive tool allows you to identify the steps to evaluate an expression using the order of operations. Printable activities, learning solutions, and other math games are available at this site.

• Weidner, John, for Study Stack. “Flashcards for Evaluating Expressions.”

www.studystack.com/flashcard-164919

Online flashcards enable students to practice evaluating expressions given specific values for the variables. The cards can be flipped as often as needed and reshuffled.



Lesson 1 Answer KeyLesson Pre-Assessment, p. 5 1. 79 2. 9/40 3. 12 + 9.75x; $31.50 4. lw = (10 feet)(5 feet); 50 square feet 5. 2(2πr); 12π

Guided Practice, p. 21 1. expression 2. equation 3. expression 4. equation 5. –1 6. 29 7. –70 8. –66 9. 64π square inches 10. 120 cubic feet 11. 252 books 12. 30 + 45x; $300

13. A l l l=

=2

323

2 ; 96 square feet

14. 0.20x; 0.05x + 30; 20; $35 15. 2.50(5) + 2.60(5) + 2.55(5) = $38.25 16. A = 100π; A = 16π; A = 100π – 16π; 84π square inches

Independent Practice, p. 24 1. equation 2. equation 3. expression 4. expression 5. 22 6. 60 7. –12 8. –162 9. A = 16π square inches 10. P = 56 feet 11. 1,025 tickets 12. 30x + 25xy; $620

13. A b b b=

=1

214

18

2 ; A = 12.5 square feet

14. 5.99x; 9.99x + 5; $59.90; $104.90 15. 3.27(2) + (3.27 – 0.50)(2) + (1.00 + (3.27 – 0.50))(2) =

$19.62 16. A = 144 square inches; A = 81 square inches; A = 63 square

inches

Progress Assessment, p. 27 1. –22 2. –48 3. $15x + $12xy; $102 4. $25x + $20; $75x; $145; $375 5. A = 4π; 4π square inches 6. P = 3(a + b + c); P = 36 feet

Unit 2 • operations with algebraic expressionsLesson 2: Adding and Subtracting Monomials

naMe:



Assessment

Lesson Pre-AssessmentList the like terms for each polynomial.

1. 3b + 3b2 – 4b + 8 – 7

For problems 2–5, combine like terms and then rewrite the new polynomials.

2. 14x + 4 + 3xy – 9xy + 17

3. x + 3x3 – 5x + 8x3 + 44

4. A car production line produces vans, sedans, and convertibles according to the following polynomial: 10v + 4s2 + 35c + 20v + 3s + 70c.

5. The number of machine parts produced, p, and the number of defective parts produced, d, at a factory each day follows the polynomial 200p + 5p2 – 14d + 74p – 2d 2 – 4d.



Instruction

Lesson 2: Adding and Subtracting Monomials

Essential Questions 1. How are properties of real numbers related to monomials?

2. How are the operations of addition and subtraction with monomials similar to those of real numbers?

WORDS TO KNOW


exponent in the expression xn, n is the exponent and indicates the number of times x is used as a factor

like terms terms that contain the same variables raised to the same power

monomial an expression that contains only one term, such as 4x or 6bc

polynomial an expression consisting of the sum of two or more terms, such as 6x2 + 5x + 4

power the exponent of an expression

term a number or variable, or the product or quotient of numbers and variables



naMe:



Warm-Up Option 1For each polynomial, determine the like terms.

1. 2x + 10x2 – 4x + 5

2. 5ab + 5cd – 5ab2 – 4cd

3. 3x4 + 4x3 – x3 + 7 – 2x + 12

4. The revenue from selling a specific printer at Offices-R-Us can be found using the expression 200x, where x is the number of printers sold. Office Supply Discount Store sells the same number of printers as Offices-R-Us, but for a lower price of $180 each. How can you write an expression for printer sales at Office Supply Discount Store so that it is a like term to that of Offices-R-Us?

5. The cost of movie tickets is $10 for adults and $6 for children. If a large group buys movie tickets, what is the relationship between the number of adults and number of children if the expression 10x + 6x represents the cost of the group’s tickets?


naMe:


Warm-Up Option 2A weather station recorded the actual rainfall in inches for May, June, July, and August in a rural area. August was the month with the least rainfall. July had 1.5 times the amount of rainfall in August, and June had 3 times the amount of rainfall in August. May had 2 inches of rain.

1. How can you use variables to represent the rainfall of each of the four months?

2. What are the like terms?

3. If 1.25 inches of rain fell in August, what was the total rainfall for the given months?




Instruction

Warm-Up Option 1: Debrief 1. 2x + 10x2 – 4x + 5

• Like terms contain the same variables raised to the same power.

• Remind students that 10x2 and 2x are not like terms because x has an exponent of 2.

• Remind students that when looking for like terms, also look at the operation directly to the left of the coefficient in order to determine the coefficient’s sign.

Example: 2x + 10x2 – 4x + 5 is equal to 2x + 10x2 + (–4x) + 5.

The terms 2x and –4x are like terms because they both have the variable x.

2. 5ab + 5cd – 5ab2 – 4cd


• Again, point out to students that –5ab2 is not like 5ab because of the difference in exponents.

• Also, be sure students recognize it is –4cd and not 4cd.

The terms 5cd and –4cd are like terms.

3. 3x4 + 4x3 – x3 + 7 – 2x + 12


The terms 4x3 and –x3 are like terms.

The terms 7 and 12 are also like terms.



Instruction

4. The revenue from selling a specific printer at Offices-R-Us is found using 200x, where x is the number of printers sold. Office Supply Discount Store sells the same number of printers as Offices-R-Us, but for a lower price of $180 each. How can you write an expression for printer sales at Office Supply Discount Store so that it is a like term to that of Offices-R-Us?


• Point out that the term for Offices-R-Us is 200x, where x represents the number of printers sold.

• Guide students to realize that the term for Office Supply Discount Store must have the variable x to compare the same type of printer.

Therefore, 180x is the like term.

5. The cost of movie tickets is $10 for adults and $6 for children. If a large group buys movie tickets, what is the relationship between the number of adults and number of children if the expression 10x + 6x represents the cost of the group’s tickets?

• Determine if the given expression contains like terms.

10x and 6x are like terms because the variables are the same and are raised to the same power.

The number of children and adults in the group must be equal because the terms are alike.

• Work through an example to demonstrate the validity of the statement.

Suppose there were 5 children and 5 adults in the group.

The cost of the tickets would be 10(5) + 6(5) or 50 + 30 = $80.

• Demonstrate to students that if the number of adults and the number of children differed, you could not use the same variable x to represent the number of each in the group.

If there were 5 adults and 12 children in the group, the expression would be:

10x + 6y = 10(5) + 6(12) = 50 + 72 = $122




Instruction

Warm-Up Option 2: DebriefA weather station recorded the actual rainfall in inches for May, June, July, and August in a rural area. August was the month with the least rainfall. July had 1.5 times the amount of rainfall in August, and June had 3 times the amount of rainfall in August. May had 2 inches of rain.

1. How can you use variables to represent the rainfall of each of the four months?

• In the problem, July and June are both compared to August, but we do not know what the rainfall was in August.

• The easiest remedy would be to assign a variable, x, to August.

• Use this variable as a base for assigning values to the other months:

August: x

July: 1.5x

June: 3x

May: 2

May does not require a variable because we know the exact number of inches of rainfall it had (2).

2. What are the like terms?

x, 1.5x, and 3x are all like terms because they all have the same variable raised to the same power.



Instruction

3. If 1.25 inches of rain fell in August, what was the total rainfall for the given months?

• Write an expression for the sum of the four months’ rainfall.

x + 1.5x + 3x + 2

• Substitute 1.25 for the variable x.

1.25 + 1.5(1.25) + 3(1.25) + 2

• Remind students to follow the order of operations by first completing all the multiplication in the problem from left to right.

1.25 + 1.875 + 3.75 + 2

• Add the terms from left to right.

1.25 + 1.875 + 3.75 + 2 = 8.875

The total amount of rainfall for the months given was 8.875 inches.


naMe:



Focus ProblemAndrea is making cube-shaped piñatas. She first makes the frame by building the edges of the cube out of thin wire, then uses pieces of paper for the sides, and fills the piñata with shredded paper.

1. First, she makes a piñata x inches wide. Find the amounts of wire needed for the edges, paper needed for the sides, and shredded paper needed for the filling.

2. Andrea then makes a piñata twice as wide as the first one. She will need 24x inches of wire, 24x2 square inches of paper for the sides, and 8x3 cubic inches of shredded paper to fill her piñata. What are the total amounts of each material needed for both piñatas?

3. Algebraically, why do we need three amounts for the total materials? Why not add them together as one?



Instruction


To introduce like terms, write the following example on the board: 10y and –3y + 2y2.

• First point out the coefficient of the variables in each monomial.

Coefficients: 10, –3, and 2

• Point out the variables of each monomial: y.

• Point out the power of the variables in the monomial.

• The first two terms have a power of 1; the last term has a power of 2.

• It is common for students to state that 10y and –3y do not have a power. Remind students that 10y = 10y1.

Write the following example on the board: Combine 3x2 and –4x2.

• Identify the coefficients: 3 and –4

• Identify the variable(s): x

• Identify the power(s): 2

3x2 and –4x2 are like terms because both terms have the same variable base of x and the same power of 2. These two terms can be combined.

• To combine like terms, add or subtract the coefficients, as you would integers, and multiply the result by the common variable(s).

3 + (–4) = –1

• Multiply –1 by the variable x2 to get –1x2 or –x2.

3x2 + –4x2 = –x2





Combine the like terms in the following list of monomials:

–4x2, 3xy, 4, 12x2y, 10xy, –12

Solution

• Identify the like terms.

• 3xy and 10xy are like terms.

• 4 and –12 are like terms.

• Identify the coefficients of each set of like terms, add/subtract the coefficients, and multiply the result by the common variable(s).

• 3 + 10 = 13, which becomes 13xy.

• 4 – 12 = –8; there are no variables.

• Point out to students that –4x2 and 12x2y are NOT like terms because of the y variable in the second term.

Example 2

Combine the like terms in the following list of monomials:

2x2, 2xy, 2, 3x2y, 10xyz2

Solution

• Identify the like terms.

• There are no like terms because none of the monomials have the same variables raised to the same power.

The monomials in the list cannot be combined because there are no like terms.



Instruction


Andrea is making cube-shaped piñatas. She first makes the frame by building the edges of the cube out of thin wire, then uses pieces of paper for the sides, and fills the piñata with shredded paper.

Question 1

First, she makes a piñata x inches wide. Find the amounts of wire needed for the edges, paper needed for the sides, and shredded paper needed for the filling.

Instruction

To determine the amount of wire needed for the piñata, identify the lengths of each edge of the piñata.

• If the width of the cube-shaped piñata is x inches, each edge must also be x inches long.

• A cube has 12 edges.

• Students may need to see a cube to count the number of edges.

Andrea needs a total of 12x inches of wire.

The amount of paper needed for each side can be determined by finding the area of each side.

• The area of a square is A = length • width = x • x.

• The area of each side of the cube is x2.

The cube has 6 sides, so in total she will need 6x2 square inches of paper.

To determine the amount of shredded paper needed for filling, Andrea must find the volume of the cube-shaped piñata.

• The volume of a cube is Volume = length • width • height.

• Each edge has a length of x.

• The volume of the piñata is x • x • x = x3.

Andrea needs a total of x3 cubic inches of shredded paper.

Andrea needs 12x inches of wire, 6x2 square inches of paper, and x3 cubic inches of shredded paper.





Andrea then makes a piñata twice as wide as the first one. She will need 24x inches of wire, 24x2 square inches of paper for the sides, and 8x3 cubic inches of shredded paper to fill her piñata. What are the total amounts of each material needed for both piñatas?

Instruction

To find the total amounts of materials needed, we can add separately the amounts of wire, paper, and shredded paper needed for each piñata. In other words, we need to add the like terms.

Wire: 12x + 24x = 36x inches

Paper: 6x2 + 24x2 = 30x2 square inches

Shredded paper: x3 + 8x3 = 9x3 cubic inches

Question 3

Algebraically, why do we need three amounts for the total materials? Why not add them together as one?

Instruction

• Remind students that like terms not only have the same variable, but are also raised to the same exponent.

• We cannot add inches of wire and square inches of paper because they are different materials and are in different units of measurement.

• Likewise, we cannot add either of these to cubic inches of shredded paper.

• Algebraically, we cannot add 24x, 24x2, or 9x3 because they are not like terms.



Instruction


Combine like terms in the following list of monomials:

ab, –3abc, 14, –3b2c, 10a2, 12b2c, 3ab, 12.

Solution

• Make a list of like terms.

The first group of like terms is ab and 3ab.

The second group of like terms is –3b2c and 12b2c.

The third group of like terms is 14 and 12.

• For each group of like terms, add or subtract the coefficients and multiply the result by the variables.

ab + 3ab = 4ab

–3b2c + 12b2c = 9b2c

14 + 12 = 26

Example 2

Combine the like terms in the given expression and rewrite the new polynomial.

5x4 – 7y + 12 + 5y – 9x4 + x4 – 2xy – 7

Solution

• Make a list of the like terms.

The first group of like terms is 5x4, –9x4, and x4.

The second group of like terms is –7y and 5y.

The third group of like terms is 12 and –7.




Instruction


5x4 – 9x4+ x4 = –3x4

–7y + 5y = –2y

12 – 7 = 5

• Rewrite the polynomial.

–3x4 – 2y – 2xy + 5

Example 3

Combine the like terms in the given expression and rewrite the new polynomial.

3x3 + 9 + 5yz – 10x3y + 3x4 – 2xy

Solution

• Make a list of the like terms.

• There are no like terms.

Since there are no like terms to combine, the polynomial has not changed. There is no need to rewrite it.

• Point out to students that it doesn’t matter if the coefficients are the same; it is the variable and the power it is raised to that must be the same in order to combine like terms.


naMe:


Guided PracticeFor problems 1–4, identify the like terms.

1. 3x2 – 3xy + 10 + x2 – 6 3. z2y + y2z – 8

2. y + 2x2 + 7 – 8 4. xy + xy2 – 4xy2 + 5x + 8x + xy + 60

For problems 5–12, combine the like terms and rewrite the new polynomial.

5. 3c + 2cd – 4cd – 3c + 7

6. 5x4 – 3x + x4 + x2 – 6 + 10x

7. 14 + 2y + 3x2 + 8x2 – 18 + 3xy

8. a2b + 2a2b – 4 + 7a2 + 7b2 + 4

9. x + 9xy2 – 4xy2 + 7x + 9x + xy + 12 + 2

10. 2rt – 16 + 3r3 + 7 – rt + 9 – 3r3

11. 5xy + 4x2 – 7xy + 8x2

12. 3a2 + 4b + 2c2 + 6b – 4a2 + abc – 2ab

continued


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For problems 13–15, list the like terms for the monomials given in each scenario.

13. An electronics store sells x number of computers, 2x number of stereos, and y number of televisions.

14. The total revenue from the sale of x number computers is 600x. The total revenue of the sale of 2x number of stereos is 325(2x). The total revenue of the sale of x number of televisions is 1,075x.

15. The electronics store lists the number of computers sold in January, February, and March as x,

3x, and 2x. The number of printers sold was y, y, and 5y. The number of computers returned

during these three months was 116

x .

For problem 16, combine like terms in each situation and rewrite the new polynomial.

16. On Saturday, an ice cream shop sells x sugar cones for $2 each, y waffle cones for $4.50 each, and z sundaes for $3.20 each. The amount of sales of sugar cones is represented by 2x, sales of waffle cones is 4.50y, and sales of sundaes is 3.20z. On Sunday, all cones are sold at half price.

The following polynomial shows the total amount of sales from Saturday and Sunday of last week: 2x + 4.50y + 3.20z + x + 2.25y + 1.60z.


naMe:


Independent Practice For problems 1–4, identify the like terms.

1. x2 – 3x + 7 + 2x2 – 3 3. z2y + 14y + 3z2y – 8

2. 3xy + x2 + 7xy – 24 4. 2xy + 5xy – 2xy2 + 3x + xy + 10

For problems 5–13, combine the like terms and rewrite the new polynomial.

5. 2a + 2ac – 5ab – 3b + 7c

6. x4 – 3x2 + 2x4 – x2 – 8 + 9x

7. 8 + 3xy + 3x2y + 9x2 – 3 + 10xy

8. 2a2b + 4a2b – 4a + 8a2 – b2 + 6

9. xy + 7xy2 – 3xy2 + 7xy + 7 + 3xy – 25

10. 3r – 9 + 2r3 + 7r – rt + 8 – 18r3

11. 1.50a + 2.50b + 0.75a + 1.50c + 2.50b + 0.75b

12. 2a2 + 10b2 + 3c3 + 7b – a2 + abc2 – 2abc2

13. 3x2y – 2x2 – 6x2y + 8xy2 + 9x2

continued


naMe:



For problems 14–16, list the like terms for the monomials given in each scenario.

14. A clothing boutique sells s number of sweaters at $18 each, c number of coats at $42 each, h number of printed hats for $9, and h number of solid hats for $7.

15. The total profit from the sale of s number of sweaters is 55s. The total profit of the sale of c number of coats is 125c. The total profit of the sale of 3c number of hats is 19.50(3c).

16. The clothing boutique has a sale and discounts the sweaters and hats. The total number of sweaters sold during the sale was 8s. The total number of coats sold during the sale was 9c. No hats were sold during the sale.


naMe:


Assessment

Progress AssessmentList the like terms for each polynomial.

1. 2ac + 3a2 – 5ac + 8

2. 12 + 4x + 5xy – 3xy + 5

For each polynomial, combine like terms and rewrite the new polynomial.

3. 6x + 3x + 5y – 2xy2 + 4y + 2xy2

4. r2 + 5s + 12 + 9s – 5r2

Solve the following problems.

5. At the farmers’ market, Wanda buys a pounds of apples for $3 per pound, b pounds of pears for $4 per pound, and another b pounds of plums for $6 per pound. The following week at the same farmers’ market, Wanda buys c pounds of apples, a pounds of pears, and c pounds of plums, all at the same prices. What is the total amount that Wanda spends on fruit over these two weeks?

6. Pedro puts in his backpack 2 books that weigh x each, 3 notebooks that weigh y each, his lunch that weighs 5x, a jacket that weighs 3y, and his MP3 player that weighs z. What is the total weight of the contents of his backpack?




Instruction

Resource List• The Math Games.com. “Identify Like Terms Game.”

http://themathgames.com/our-games/like-terms-games/combine-like-terms-game

This interactive math game allows you to identify like terms. Sound is optional, but not required. Choose a one- or two-player game and select the difficulty level.

• Weidner, John, for Study Stack. “Combining Like Terms Flashcards.”

www.studystack.com/flashcard-47007

Online flashcards enable students to practice combining like terms. The cards can be flipped as often as needed and reshuffled.



Lesson 2 Answer KeyLesson Pre-Assessment, p. 30 1. 3b and –4b; 8 and –7 2. –6xy + 14x + 21 3. 11x3 – 4x + 44 4. 4s2 + 30v + 105c + 3s 5. 5p2 – 2d2 + 274p – 18d

Guided Practice, p. 45 1. 3x2 and x2; 10 and –6 2. 7 and –8 3. no like terms 4. xy and xy; xy2 and –4xy2; 5x and 8x 5. –2cd + 7 6. 6x4 + x2 + 7x – 6 7. 11x2 +3xy + 2y – 4 8. 3a2b + 7a2 + 7b2 9. 5xy2 + xy + 17x + 14 10. rt 11. 12x2 – 2xy 12. –a2 + 2c2 + 10b + abc – 2ab 13. x and 2x 14. 600x, 325(2x), and 1,075x 15. x, 3x, 2x, and 1/16x; y, y, and 5y 16. 3x + 6.75y + 4.8z

Independent Practice, p. 47 1. x2 and 2x2; 7 and –3 2. 3xy and 7xy 3. z2y and 3z2y 4. 2xy, 5xy, and xy 5. 2a + 2ac – 5ab – 3b + 7c 6. 3x4 –4x2 + 9x – 8 7. 3x2y + 9x2 + 13xy + 5 8. 6a2b + 8a2 – b2 – 4a + 6 9. 4xy2 + 11xy – 18 10. –16r3 – rt + 10r – 1 11. 2.25a + 5.75b + 1.50c 12. 3c3 – abc2 + a2 + 10b2 +7b 13. –3x2y + 7x2 + 8xy2

14. 9h and 7h 15. 125c and 19.50(3c) 16. no like terms

Progress Assessment, p. 49 1. 2ac and –5ac 2. 5xy and –3xy; 12 and 5 3. 9x + 9y 4. –4r2 + 14s + 12 5. 7a + 10b + 9c 6. 7x + 6y + z

Unit 2 • operations with algebraic expressionsLesson 3: Adding Polynomials

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Assessment

Lesson Pre-AssessmentAdd the polynomials.

1. (2x2 – 5xy) + (4xy + 3x2 – 2)

2. (x2 + 10y) + (10x + 3x2 – 6y)

3. (a2 + abc 2 + 3) + (7ab +2a2)

4. An engineer determined that the production of a toy factory is (x2 – 5x + 2) toys, and that the production of its subsidiary factory is (3x2 – 12) toys. What is the total toy production of the toy factory and its subsidiary?

5. In the month of November, the total energy consumed by a house is determined by the polynomial (3 + 2t2 – 4s). In December, the same house consumes (2s2 + 6s + 7). What is the total energy consumption of the house in November and December altogether?



Instruction

Lesson 3: Adding PolynomialsEssential Questions 1. Why are like terms necessary when adding polynomials?

2. How is addition of polynomials similar to that of real numbers?

WORDS TO KNOW








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Warm-Up Option 1For each polynomial, determine the like terms.

1. 3x + 4x3 – 2x + 2 + 5x3 – 5

2. 3xy + 2yz – 3xy2 – xy + 3

3. 2x6 + 3x6 – 3x3 + 7x – 8x + 15

4. The number of computers a technology company produces in one week follows the polynomial 2x3 + 3x + 4x3 + 9x + 25x. Determine the like terms in this scenario.

5. A pizza parlor has determined that the average number of slices ordered at lunch is 2x, where x is the number of customers at lunch. The average number of breadsticks ordered is 3x, and the average number of drinks ordered is 1x. Why are these three expressions like terms?


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Warm-Up Option 2Allen’s Car Dealership sells a specific sedan for $21,000. Bob’s Discount Cars sells the same sedan for $19,995. Costa’s Cars sells the same sedan for $22,345. Allen’s and Bob’s sell the same convertible for $17,990 and $18,190 respectively. The number of sedans sold at each dealership is defined by the variable s, and the number of convertibles sold at each dealership is defined by the variable c.

1. How would you determine each dealership’s revenue for each type of car sold?

2. What are the like terms in the problem?

3. How do you combine like terms?

4. What new terms result when you combine the like terms?




Instruction

Warm-Up Option 1: Debrief 1. 3x + 4x3 – 2x + 2 + 5x3 – 5

• Remind students that like terms contain the same variables raised to the same power.

• Remind students that when looking for like terms, to also look at the operation directly to the left of the coefficient in order to determine the coefficient’s sign.

3x + 4x3 – 2x + 2 + 5x3 – 5 is equal to 3x + 4x3 + (–2x) + 2 + 5x3 + (–5)

• The terms 3x and –2x are like terms because both contain the variable x raised to the power of 1.

• The terms 4x3 and 5x3 are like terms because both contain the variable x raised to the power of 3.

• The terms 2 and –5 are like terms because both terms are constants.

2. 3xy + 2yz – 3xy2 – xy + 3


• The terms 3xy and –xy are like terms because both terms contain the variables x and y raised to the power of 1.

• It is common for students to list –3xy2 with 3xy and –xy because it also contains the variables xy. Remind students that exponents must also match.

3. 2x6 + 3x6 – 3x3 + 7x – 8x + 15

• The terms 2x6 and 3x6 are like terms because both terms contain x raised to the power of 6.

• The terms 7x and –8x are also like terms as they both contain the variable x.

4. The number of computers a technology company produces in one week follows the polynomial 2x3 + 3x + 4x3 + 9x + 25x.

• The terms 2x3 and 4x3 are like terms as they both contain the variable x raised to the third power.

• The terms 3x, 9x, and 25x are also like terms as they all contain the variable x.



Instruction

5. A pizza parlor has determined that the average number of slices ordered at lunch is 2x, where x is the number of customers at lunch. The average number of breadsticks ordered is 3x. The average number of drinks ordered is 1x. Why are these three expressions like terms?

• 2x, x, and 3x are like terms because they have the same variable x raised to the power of 1.

• In this case, x represents the number of customers, which can be determined for any lunch.

Warm-Up Option 2: DebriefAllen’s Car Dealership sells a specific sedan for $21,000. Bob’s Discount Cars sells the same sedan for $19,995. Costa’s Cars sells the same sedan for $22,345. Allen’s and Bob’s sell the same convertible for $17,990 and $18,190 respectively. The number of sedans sold at each dealership is defined by the variable s, and the number of convertibles sold at each dealership is defined by the variable c.

1. How would you determine each dealership’s revenue for each type of car sold?

• To determine revenue, multiply the cost of each car by the number of cars sold.

Allen’s Car Dealership: 21,000s and 17,990c

Bob’s Discount Cars: 19,995s and 18,190c

Costa’s Cars: 22,345s

2. What are the like terms in the problem?

• One group of like terms is 21,000s, 19,995s, and 22,345s.

• The second group of like terms is 17,990c and 18,190c.

3. How do you combine like terms?

• To combine like terms, add/subtract the coefficients, as you would integers, and multiply the sum by the common variables.

4. What new terms result when you combine the like terms?

• The terms are positive, so add them.

21,000s + 19,995s + 22,345s = 63,340s

17,990c + 18,190c = 36,180c


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Focus ProblemA marketing research institute is evaluating the impact of a new advertisement campaign on the profits of a certain chain of stores in 6 cities. For the first month of the campaign, the stores’ profits and losses are represented by the polynomial 4x2 – 6x + 7 + 3x2 – 5 + 7x4. Profits and losses for the second month are represented by the polynomial x2 – 3x + 2 + 3x4 + 3x6.

1. What were the total profits and losses in the first month of the campaign?

2. What were the total profits and losses in the second month of the campaign?

3. What were the total profits and losses over the two months?

4. Profits and losses in the third month of the campaign are exactly the same as the second month. What are the total profits and losses over the three months?



Instruction


To introduce like terms, write the following example on the board:

Polynomial 1: 3y + 4 – 8y + y2

Polynomial 2: 5 + 6x + 5y2

• Identify the coefficients of each polynomial.

Polynomial 1: 3, 4, –8, and 1

Polynomial 2: 5, 6, and 5

• Identify the variables of each polynomial.

Polynomial 1: y and y2

Polynomial 2: x and y2

• Identify the power of the variables of each polynomial.

Polynomial 1: 1, 1, and 2

Polynomial 2: 1 and 2

• Like terms are terms with the same variable raised to the same coefficient.

The like terms of both polynomials are 3y and –8y; 4 and 5; y2 and 5y2.

To introduce how to combine like terms, write the following example on the board:

Combine like terms from the polynomials 3y + 4 – 8y + y2 and 5 + 6x + 5y2.

• Identify the like terms; it is helpful to use a different pen color to highlight different terms.

• 3y and –8y (Be sure students correctly identify the term as –8y rather than 8y)

• 4 and 5

• y2 and 5y2




Instruction

• Combine like terms by adding or subtracting the coefficients, as you would integers, and multiply the result by the common variable(s).

• The sum of the coefficients for like terms 3y and –8y are 3 + (–8), which equals –5.

• Multiply –5 by the variable y to get –5y.

• The sum of the coefficients for like terms 4 and 5 is 4 + 5, which equals 9.

• The sum of the coefficients for like terms y2 and 5y2 is 1 + 5 = 6.

• Multiply 6 by the variable y2 to get 6y2.

• To add polynomials, group their like terms together and find their sum.

• The sum of the two polynomials includes like terms that have been combined and also terms that didn’t have any like terms.

(3y + 4 – 8y + y2) + (5 + 6x + 5y2) = 6y2 – 5y + 6x + 9

• Explain to students that the polynomials being added do NOT need to have the same number of terms.

Example 1

Add the polynomials: (2x + 4x2 + 6xy – 12x) + (2xy – 5x2 + 10xy – 20).

Solution

• Identify like terms.

(2x + 4x2 + 6xy – 12x) + (2xy – 5x2 + 10xy – 20)

• The first group of like terms is 2x and –12x, which combine to –10x.

(–10x + 4x2 + 6xy) + (2xy – 5x2 + 10xy – 20) = (–10x – x2 + 6xy) + (2xy + 10xy – 20)

• The second group of like terms is 4x2 and –5x2, which combine to –x2.

(–10x – x2 + 6xy) + (2xy + 10xy – 20)

• The third group of like terms is 6xy, 2xy, and 10xy, which combine to 18xy.

(–10x – x2 + 18xy) – 20



Instruction

• Rewrite the polynomial with the combined monomials and the leftover monomials.

–10x – x2 + 18xy – 20

Example 2

Add the polynomials: (r3 + 3t2 + 4r3 – 15) + (2t – 5s2 – 25 + 3r3).

Solution


(r3 + 3t2 + 4r3 – 15) + (2t – 5s2 – 25 + 3r3) = 8r3 + 3t2 – 15 + 2t – 5s2 – 25

• The first group of like terms is r3, 4r3, and 3r3, which combine to 8r3.

8r3 + 3t2 – 15 + 2t – 5s2 – 25 = 8r3 + 3t2 + 2t – 5s2 – 40

• The second group of like terms is –15 and –25, which combine to –40.

• Rewrite the polynomial with the combined monomials and the leftover monomials.

8r3 + 3t2 + 2t – 5s2 – 40


A marketing research institute is evaluating the impact of a new advertisement campaign on the profits of a certain chain of stores in 6 cities. For the first month of the campaign, the stores’ profits and losses are represented by the polynomial 4x2 – 6x + 7 + 3x2 – 5 + 7x4. Profits and losses for the second month are represented by the polynomial x2 – 3x + 2 + 3x4 + 3x6.





What were the total profits and losses in the first month of the campaign?

Instruction

The polynomial 4x2 – 6x + 7 + 3x2 – 5 + 7x4 represents profits and losses in the six cities.

• To add these amounts and find the total profits and losses for the month, first find like terms.

4x2 and 3x2

7 and –5

• Group like terms to add.

4x2 + 3x2 + 7 – 5 – 6xy + 7x4

• Add like terms.

7x2 + 2 – 6x + 7x4

The total profits and losses in the first month were 7x2 + 2 – 6x + 7x4.

Question 2

What were the total profits and losses in the second month of the campaign?

Instruction

The polynomial x2 – 3x + 2 + 3x4 + 3x6 represents profits and losses in the six cities.

• To add these amounts and find the total profits and losses for the month, find like terms.

• There are no like terms within this polynomial.

• The polynomial cannot be simplified.

The total profits and losses in the second month were x2 – 3x + 2 + 3x4 + 3x6.




What were the total profits and losses over the two months?

Instruction

• To add the polynomials for the two months, find their like terms.

• Remember that we have simplified at least one of them by combining like terms.

• The two polynomials are:

First month: 7x2 + 2 – 6x + 7x4

Second month: x2 – 3x + 2 + 3x4 + 3x6

• Identify like terms in both polynomials.

7x2 and x2

–6x and –3x

2 and 2

7x4 and 3x4

• Remind students that polynomials are added by grouping their like terms together and finding their sum.

• Combine all the like terms between the polynomials. The result will include the combined terms as well as the leftover terms.

(7x2 + 2 – 6x + 7x4) + (x2 – 3x + 2 + 3x4 + 3x6)

= 7x2 + x2 + 2 + 2 – 6x – 3x + 7x4 + 3x4 + 3x6

= 8x2 + 4 – 9x + 10x4 + 3x6

• Rewrite the polynomial in descending order. Arrange the terms so that the term with the highest power is listed first. Continue to the next highest power.

= 3x6 + 10x4 + 8x2 – 9x + 4

The total profits and losses over the two months were 3x6 + 10x4 + 8x2 – 9x + 4.





Profits and losses for the third month of the campaign are exactly the same as the second month. What are the total profits and losses over the three months?

Instruction

• To add the polynomials, find the like terms of the third month and the sum of the first two months found in the previous question.

• First two months: 3x6 + 10x4 + 8x2 – 9x + 4

• Third month: x2 – 3x + 2 + 3x4 + 3x6

• Identify like terms in both polynomials.

8x2 and x2

–9x and –3x

4 and 2

10x4 and 3x4

3x6 and 3x6

• Combine all the like terms between the polynomials. The result will include the combined terms as well as the leftover terms.

(3x6 + 10x4 + 8x2 – 9x + 4) + (x2 – 3x + 2 + 3x4 + 3x6)

= 8x2 + x2 – 9x – 3x + 10x4 + 3x4 + 3x6 + 3x6 + 4 + 2

= 9x2 – 12x + 13x4 + 6x6 + 6

• Rewrite the polynomial in descending order.

6x6 + 13x4 + 9x2 – 12x + 6

The total profits and losses over the three months were 6x6 + 13x4 + 9x2 – 12x + 6.



Instruction


Add the polynomials: (2t2 – 2r4 – 15t2 + 19) + (12 – 3s4 – 8t2).

Solution

• Make a list of like terms in both polynomials.

2t2, –15t2, and –8t2

19 and 12


2t2 – 15t2 – 8t2 = –21t2

19 + 12 = 31

• Write the result including the combined terms and leftover terms.

–2r4 – 3s4 – 21t2 + 31

Example 2

Add the polynomials: (14 + x2 – 3xy4 – 4x3 + 24x) + (4x3 – 2xy4 – 9).

Solution

• Make a list of like terms in both polynomials.

14 and –9

–3xy4 and –2xy4

–4x3 and 4x3




Instruction


14 – 9 = 5

–3xy4 + (–2xy4) = –5xy4

–4x3 + 4x3 = 0

• Write the result including the combined terms and leftover terms in descending order.

–5xy4 + x2 + 24x + 5

Example 3

Add the polynomials: (3x3 + 5) + (9 + 3xy – 10x3y) + (2x3 – 6x3y).

Solution

• Make a list of like terms in the polynomials.

3x3 and 2x3

5 and 9

–10x3y and –6x3y


3x3 + 2x3 = 5x3

5 + 9 = 14

–10x3y + (–6x3y) = –16x3y

• Write the result including the combined terms and left over terms in descending order.

–16x3y + 5x3 + 3xy + 14


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Guided PracticeAdd the polynomials.

1. (x2 – 6xy) + (10xy +5x2 – 6)

2. (x2 + 4) + (7y + 5x2 – 8)

3. (b3 + 10) + (6b + 8f 2 + 8)

4. (y5 + 3xy – 8xy ) + (xy + y5 – 10xy)

5. (ab2 + ab + ab) + (4abc – 4a2 – 21)

6. (x4 – 2x) + (4x4 + x) + (15 – 9x)

7. (5x2 – x2) + (2x4 + y) + (14x – 10y)

8. (14 + 3c4 – 3c) + (c4 + 12 + d) + (5c4 – 10d + 2)

9. (2s4 – s) + (t 4 + s) + (6t 5 – 10s + 12t 4)

10. (s2 – s2 + 12) + (2t 4 + s) + (5s 5 – 9 + 7t 5 + 50)

continued


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For problems 11–16, add the polynomials given in each scenario.

11. A statistician used the mathematical polynomials (x3 – 6x) and (12y + 14x2 – 17) to determine the odds of winning a card game. What is the sum of the polynomials?

12. Jacquie has to record the amounts of oxygen produced by two different plants for her science project. The amount of oxygen produced by one plant is represented by the polynomial (12 + t 2 – 6s). The other plant’s oxygen is represented by the polynomial (s2 – 6s). How much total oxygen did the two plants produce?

13. Two competing production lines at a factory produced heating and air conditioning parts following the polynomials (12a + 4a2) and (a2 – 6 + 14). What is the sum of the polynomials?

14. The profit for three different companies is represented by the polynomials (3y2 + 4), (y2 – 8), and (10y2 – 1). What is the sum of the polynomials?

15. The flight of a bird can be represented by a polynomial with respect to the ground. One bird flies according to the polynomial (–2r2 – 6r + 5). Another bird flies according to the polynomial (r2 + 10r + 14). What is the sum of the birds’ flight paths?

16. The swim path of a fish can be represented by a polynomial with respect to the ocean floor. Two fish swim together according to the polynomial (xy2 – 4y + 2). What is the sum of the two fish paths?


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Independent Practice Add the polynomials.

1. (2x2 – 5xy2) + (xy + 4x2 – 5)

2. (–3x2 + 4y) + (7y + 9x2 – 24)

3. (3b3 + 10f 2) + (b – 8f 2 – 4)

4. (y5 – 2xy – xy ) + (5xy – 3y5 – 10y)

5. (b2 + abc + ab) + (2ab – 2a2 + 16)

6. (2x4 – 5x) + (3x + x) + (15y – 11)

7. (x2 – 3x2) + (x4 – 10y) + (5x + 6y)

8. (14c4 – 2c4 – 3c) + (2c4 – 18 + d2) + (c4 – 4d2 – 1)

9. (s4 – s) + (st 4 + s) + (6st 5 – 4s + 19t 4)

10. (5s2 – s2 + 7t 4) + (11t 4 + s) + (4s5 + 9 – 7t 5 – 26)

continued


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For problems 11–16, add the polynomials given in each scenario.

11. A scientist used the mathematical polynomials (2x3 – x) and (12y –2x2 + 18) to determine the odds of a correct hypothesis. What is the sum of the polynomials?

12. A campaign manager derived polynomials based on voter responses to campaign polls. The polynomials were (2t 2 – 4s + 10) and (5t 2 – 2s + 7). What is the sum of the polynomials?

13. Over the past two days, a production line at a factory produced hammers according to the following polynomials: (5f + 4f 3) and (20f 3 – 5f – 2). What is the sum of the polynomials?

14. The number of books sold by three different bookstores is represented by the polynomials (2y2 + 4y), (3y2 – 3), and (y2 – y). What is the sum of the polynomials?

15. The flight of a mosquito can be represented by a polynomial with respect to the ground. One mosquito flies according to the polynomial (t 2 – 8t + 10). Another mosquito flies according to the polynomial (t 2 + 12). What is the sum of the mosquitoes’ flight paths?

16. The walking path of a dog looking for a hidden treat in a maze can be represented by a polynomial with respect to the starting and ending points. The first time the dog attempts the maze, he finds the treat according to the polynomial (2y2 + 9x + 5). The second time he follows the same path. What is the sum of the polynomials of the dog’s two trips through the maze?


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Assessment

Progress AssessmentAdd the polynomials.

1. (ab + 2a2) + (–5ab + 14)

2. (5 + 2x + 10xy) + (–2xy – 1)

3. (2r + 5r2) + (4r – 5t 3 + 12)

4. (x + 5x2) + (–5x + 17y) + (10y + 3)

5. (6x + 3x + 5y) + (2xy2 + 4y) + (2xy2 – 5y)

6. A computer factory produces (r2 + 5s + 12) computers each month. A second factory with different machinery produces (9s – 5r2 – 5t) each month, and a third one produces (2r2 + 5r2 + 13) per month. In one month, how many computers do the three factories produce altogether?




Instruction

Resource List• Roberts, Donna, for Oswego City (NY) School District Regents Exam Prep Center. “Adding

Polynomials.”

www.regentsprep.org/regents/math/ALGEBRA/AV2/sp_add.htm

This Web site shows students how to add polynomials with four different methods: using colors for horizontal addition, using vertical addition, using algebra tiles, and using the distributive property to factor coefficients.

• Stapel, Elizabeth, for Purplemath. “Adding Polynomials.”

www.purplemath.com/modules/polyadd.htm

This Web site offers five detailed examples on how to add polynomials by combining like terms.



Lesson 3 Answer KeyLesson Pre-Assessment, p. 52 1. 5x2 – xy – 2 4. 4x2 – 5x – 10 2. 4x2 + 4y + 10x 5. 2t 2 + 2s2 + 2s + 10 3. 3a2 + abc2 + 7ab + 3

Guided Practice, p. 67 1. 6x2 + 4xy – 6 2. 6x2 + 7y – 4 3. b3 + 6b + 8f 2+ 18 4. 2y5 – 14xy 5. –4a2 + ab2 + 2ab + 4abc – 21 6. 5x4 – 10x + 15 7. 2x4 + 4x2 + 14x – 9y 8. 9c4 – 3c – 9d + 28 9. 2s4 – 10s + 6/5 + 13t4

10. 5s5 + s + 7t 5 + 2t 4 + 53 11. x3 + 14x2 – 6x + 12y – 17 12. –12s + s2 +t 2 + 12 13. 5a2 + 12a + 8 14. 14y2 – 5 15. –r2 + 4r + 19 16. 2xy2 – 8y + 4

Independent Practice, p. 69 1. 6x2 + xy – 5xy2 – 5 2. 6x2 + 11y – 24 3. 3b3 + b + 2f 2 – 4 4. –2y5 + 3xy – 10y 5. – 2a2 + abc + 3ab + b2 + 16 6. 2x4 – x + 15y – 11 7. x4 – 2x2 – 4y + 5x 8. 15c4 – 3c – 3d2 – 19 9. s4 – 4s + 6st 5 + st 4 + 19t 4

10. 4s5 + 4s2 + s – 7t 5 + 18t 4 – 17 11. 2x3 – 2x2 – x + 12y + 18 12. – 6s + 7t 2 + 17 13. 24f 3– 2 14. 6y2 + 3y – 3 15. 2t 2 – 8t + 22 16. 4y2 + 18x + 10

Progress Assessment, p. 71 1. 2a2 – 4ab + 14 4. 5x2 – 4x + 27y + 3 2. 2x + 8xy + 4 5. 9x + 4xy2 + 4y 3. 5r2 + 6r – 5t 3 + 12 6. 3r 2 + 14s – 5t + 25

Unit 2 • operations with algebraic expressionsLesson 4: Subtracting Polynomials

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Assessment

Lesson Pre-AssessmentSubtract the following polynomials.

1. (8y + 3x + 5) – (2xy + 3x + 2y)

2. (x2 + 8x + 7) – (–x2 – 6x – 5)

3. (3y3 + 4y2 + 5y) – (3y3 + 2y2 + y)

4. From 2002 through 2009, the number of entertainment video games, E, and educational video games, D, can be modeled by the following equations, where t is the number of years since 2002:

Entertainment games: E = 0.478t 2 – 1.075t + 0.621

Educational games: D = 0.045t 2 – 1.032t + 0.345

Find the difference in the number of entertainment and educational games.

5. From 2000 to 2007, the number of births, B, in the United States can be modeled by the equation B = 37x + 4,059, and the number of deaths, D, can be modeled by the equation D = 3x + 2,404, where x is the number of years since 2000 and B and D are in thousands. Write an expression that can be used to model the difference in the number of births and deaths in the United States from 2000 to 2007.



Instruction

Lesson 4: Subtracting PolynomialsEssential Questions 1. Why are like terms necessary when subtracting polynomials?

2. Why do we use the additive inverse to subtract polynomials?

3. How is subtraction of polynomials similar to that of real numbers?

WORDS TO KNOW

additive inverse of a monomial the original term but with the opposite sign, such that when added to the original term the sum equals zero

additive inverse of a polynomial the original polynomial with each of its terms replaced with their additive inverses








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Warm-Up Option 1Find the sum of each of the following polynomials.

1. (6x2 + 3) + (8 + 12x2)

2. (12x5 – 9x – 7x2) + (10x + 3x5 – 4x2)

3. (–3m3 + 2n2 – 3) + (7 + 12n2 – 8m3)

Find the additive inverse of each of the following polynomials.

4. 15y – 8xy + 4x

5. x2 – 2xy + y2


naMe:


Warm-Up Option 2On Linda’s farm, each tomato plant gives her 24 tomatoes, each red bell pepper plant gives her 15 peppers, and each butternut squash plant gives her 8 squashes during the whole season. She decides that this season she will grow t tomato plants, p red bell pepper plants, and s butternut squash plants.

1. How would you determine the number of tomatoes that Linda’s farm produces throughout the season? How about the number of peppers and the number of squashes?

2. Write an expression for the total production of Linda’s farm.

3. Find the additive inverse of the polynomial that describes the total production of Linda’s farm.




Instruction

Warm-Up Option 1: Debrief 1. (6x2 + 3) + (8 + 12x2)


6x2 and 12x2

3 and 8

• Combine like terms.

6x2 + 12x2 = 18x2

3 + 8 = 11

• Rewrite the polynomial with the combined monomials and any leftover monomials.

(6x2 + 3) + (8 + 12x2) = 18x2 + 11

2. (12x5 – 9x – 7x2) + (10x + 3x5 – 4x2)


12x5 and 3x5

–9x and 10x

–7x2 and –4x2


12x5 + 3x5 = 15x5

–9x + 10x = x

–7x2 + (–4x2) = –11x2


(12x5 – 9x – 7x2) + (10x + 3x5 – 4x2) = 5x5 – 11x2 + x



Instruction

3. (–3m3 + 2n2 – 3) + (7 + 12n2 – 8m3)


–3m3 and –8m3

2n2 and 12n2

–3 and 7


–3m3 + (–8m3) = –11m3

2n2 + 12n2 = 14m2

–3 + 7 = 4


(–3m3 + 2n2 – 3) + (7 + 12n2 – 8m3) = –11m3 + 14m2 + 4

4. Find the additive inverse: 15y – 8xy + 4x

• To find the additive inverse of the polynomial, first find the additive inverse of each term.

• To find the additive inverse of a term, write the term with the opposite sign.

The additive inverse of 15y is –15y.

The additive inverse of –8xy is 8xy.

The additive inverse of 4x is –4x.

The additive inverse of 15y – 8xy + 4x is –15y + 8xy – 4x.

• It is common for students to forget that the sign of 15y is positive. Some students may fail to notice that its additive inverse is –15y.

• Remind students that when the first term of a polynomial has no written sign, it is positive and its additive inverse should be negative.




Instruction

5. Find the additive inverse: x2 – 2xy + y2

• To find the additive inverse of the polynomial, first find the additive inverse of each term.

• To find the additive inverse of a term, write the term with the opposite sign.

The additive inverse of x2 is –x2.

The additive inverse of –2xy is 2xy.

The additive inverse of y2 is –y2.

The additive inverse of x2 – 2xy + y2 is –x2 + 2xy – y2.

Warm-Up Option 2: DebriefOn Linda’s farm, each tomato plant gives her 24 tomatoes, each red bell pepper plant gives her 15 peppers, and each butternut squash plant gives her 8 squashes during the whole season. She decides that this season she will grow t tomato plants, p red bell pepper plants, and s butternut squash plants.

1. How would you determine the number of tomatoes that Linda’s farm produces throughout the season? How about the number of peppers and the number of squashes?

• To find the number of tomatoes that Linda’s farm produces, multiply the number of tomato plants by the number of tomatoes that each of those plants gives: 24t

• To find the number of peppers that Linda’s farm produces, multiply the number of red bell pepper plants by the number of peppers that each of those plants gives: 15p

• To find the number of squashes that Linda’s farm produces, multiply the number of butternut squash plants by the number of squashes that each of those plants gives: 8s

2. Write an expression for the total production of Linda’s farm.

• To find the total production of Linda’s farm, add the production of tomatoes, peppers, and squashes.

24r + 15p + 8s



Instruction

3. Find the additive inverse of the polynomial that describes the total production of Linda’s farm.

• To find the additive inverse of the polynomial, find the additive inverse of its terms.

The additive inverse of 24r is –24r.

The additive inverse of 15p is –15p.

The additive inverse of 8s is –8s.

Therefore, the additive inverse of 24r + 15p + 8s is –24r – 15p – 8s.


naMe:



Focus ProblemKym and Caroline are starting a bike and skate shop. They can buy their products at wholesale prices: road bikes for $400 each, mountain bikes for $520 each, and skateboards for $55 each. They are planning to sell those to their customers at a higher price: road bikes at $549 each, mountain bikes at $699 each, and skateboards at $79 each.

1. What is their cost of purchasing r road bikes, m mountain bikes, and s skateboards?

2. What is the revenue from selling r road bikes, m mountain bikes, and s skateboards?

3. What is their profit from purchasing and selling r road bikes, m mountain bikes, and s skateboards?

4. Kym and Caroline decide to sell their products at a discount to an outdoor adventure club for the following prices: road bikes for $440, mountain bikes for $560, and skateboards for $63. How much profit is lost by giving a discount?



Instruction


In order to subtract a polynomial B from another polynomial A, we must find the additive inverse of polynomial B and add it to polynomial A. The result will be the difference of polynomials, A – B.

• First, determine which is the polynomial to be subtracted.

• Find the additive inverse of the polynomial to be subtracted.

• Take the additive inverse found in the previous step and add it to the polynomial from which you have to subtract.

• Tell students that subtracting polynomials is not that different from subtracting numbers. Remind students of the subtraction 4 – 3 = 4 + (–3), where instead of subtracting the number 3, we add its additive inverse, –3. This is precisely the method that must be followed to subtract polynomials.

Example

Subtract the polynomials (7x + 6) – (3x + 5).

Solution

• Find the additive inverse of the polynomial to be subtracted, 3x + 5.

• The additive inverse of 3x + 5 is –3x – 5.

• Rewrite the subtraction problem as an addition problem.

(7x + 6) + (–3x – 5)

• Add the polynomials. Remember that to add polynomials you must add like terms.

The sum of like terms 7x and –3x is 4x.

The sum of like terms 6 and –5 is 1.

• Combine these terms to find the answer: 4x + 1

Therefore, (7x + 6) – (3x + 5) = 4x + 1.




Instruction


Kym and Caroline are starting a bike and skate shop. They can buy their products at wholesale prices: road bikes for $400 each, mountain bikes for $520 each, and skateboards for $55 each. They are planning to sell those to their customers at a higher price: road bikes at $549 each, mountain bikes at $699 each, and skateboards at $79 each.

Question 1

What is their cost of purchasing r road bikes, m mountain bikes, and s skateboards?

Instruction

Using the information given in the problem, you can write a polynomial that represents the cost of purchase by finding first the cost for each product and then the total cost.

• Kym and Caroline purchase r road bikes for $400 each, so they spend 400r on road bikes.

• They purchase m mountain bikes for $520 each, so they spend 520m on mountain bikes.

• They purchase s skateboards for $55 each, so they spend 55s on skateboards.

The total cost of purchasing all three products is 400r + 520m + 55s.

Question 2

What is the revenue from selling r road bikes, m mountain bikes, and s skateboards?

Instruction

Using the information given in the problem, you can write a polynomial that represents the sales revenue by finding first the revenue for each product and then the total revenue.

• Kym and Caroline sell r road bikes for $549 each, so the revenue is 549r for road bikes.

• They sell m mountain bikes for $699 each, so the revenue is 699m for mountain bikes.

• They sell s skateboards for $79 each, so the revenue is 79s for skateboards.

The total sales revenue for all three products is 549r + 699m + 79s.




What is their profit from purchasing and selling r road bikes, m mountain bikes, and s skateboards?

Instruction

To find the profit, subtract the purchasing cost from the sales revenue.

• Kym and Caroline have sales revenue of 549r + 699m + 79s.

• Their purchasing cost is 400r + 520m + 55s.

• Therefore, the profit is determined by (549r + 699m + 79s) – (400r + 520m + 55s).

To perform the subtraction, first find the additive inverse of the polynomial to be subtracted.

• The additive inverse of (400r + 520m + 55s) is (–400r – 520m – 55s).

Now set up the subtraction problem as an addition problem using the additive inverse.

(549r + 699m + 79s) + (–400r – 520m – 55s)

To add the polynomials, find the sums of the like terms:

549r + (–400r) = 149r

699m + (–520m) = 179m

79s + (–55s) = 24s

The profit is (549r + 699m + 79s) – (400r + 520m + 55s) = 149r + 179m + 24s.

Question 4

Kym and Caroline decide to sell their products at a discount to an outdoor adventure club for the following prices: road bikes for $440, mountain bikes for $560, and skateboards for $63. How much profit is lost by giving a discount?

Instruction

To find the profit lost, first write a polynomial to represent the discount sales revenue.

• Kym and Caroline sell r road bikes for $440 each, so the revenue is 440r for road bikes.

• They sell m mountain bikes for $560 each, so the revenue is 560m for mountain bikes.




Instruction

• They sell s skateboards for $63 each, so the revenue is 63s for skateboards.

• The total sales revenue for all three products is 440r + 560m + 63s.

Now subtract the purchase cost from the discount sales revenue.

• Kym and Caroline have sales revenue of 440r + 560m + 63s.

• Their purchase cost is 400r + 520m + 55s.

• Therefore, the profit is determined by (440r + 560m + 63s) – (400r + 520m + 55s).

To perform the subtraction, you must first find the additive inverse of the polynomial to be subtracted.

• The additive inverse of (400r + 520m + 55s) is (–400r – 520m – 55s).

Now set up the subtraction problem as an addition problem using the additive inverse.

(440r + 560m + 63s) + (–400r – 520m – 55s)

To add the polynomials, find the sums of the like terms:

440r + (–400r) = 40r

560m + (–520m) = 40m

63s + (–55s) = 8s

The profit on the discounted sales is 40r + 40m + 8s.

To find the amount of profit lost, subtract the discounted sales profit from the normal sales profit.

• The normal sales profit is 149r + 179m + 24s.

• The discounted sales profit is 40r + 40m + 8s.

• Therefore, the profit lost is determined by (149r + 179m + 24s) – (40r + 40m + 8s).

Now create an addition problem using the additive inverse.

149r + 179m + 24s + (–40r – 40m – 8s)



Instruction

To add the polynomials, find the sums of like terms:

149r + (–40r) = 109r

179m + (–40m) = 139m

24s + (–8s) = 16s

The amount of profit lost by giving a discount is 109r + 139m + 16s.


Subtract the polynomials: (3x2 – 12y2 + 3z2) – (5x2 – 20y2 – xyz)

Solution

• To perform the subtraction of polynomials, first find the additive inverse of the polynomial to be subtracted.

• The additive inverse of 5x2 – 20y2 – xyz is –5x2 + 20y2 + xyz.

• Rewrite the problem as an addition of polynomials.

(3x2 – 12y2 + 3z2) + (–5x2 + 20y2 + xyz)

• Add the like terms to obtain the result.

3x2 + (–5x2) = –2x2

–12y2 + 20y2 = 8y2

3z2 has no like terms.

xyz has no like terms.

(3x2 – 12y2 + 3z2) – (5x2 – 20y2 – xyz) = –2x2 + 8y2 + 3z2 + xyz





Subtract the polynomials: (2a – a3 + 6) – (2a3 + 3a2 – 7a – 3)

Solution

• To perform the subtraction of polynomials, first find the additive inverse of the polynomial to be subtracted.

• The additive inverse of 2a3 + 3a2 – 7a – 3 is –2a3 – 3a2 + 7a + 3.

• Rewrite the problem as an addition of polynomials.

(2a – a3 + 6) + (–2a3 – 3a2 + 7a + 3)

• Add the like terms to obtain the result.

–a3 + (–2a3) = –3a3

–3a2 has no like terms.

2a + 7a = 9a

6 + 3 = 9

(2a – a3 + 6) – (2a3 + 3a2 – 7a – 3) = –3a3 – 3a2 + 9a + 9


naMe:


Guided PracticeSubtract the following polynomials.

1. (–80a + 36b – 12ab) – (16a + 40b + 20ab)

2. (6p – 26q – 14r) – (–12p – 4q + 11r)

3. (–3g + 7h – 5) – (3g + 7h + 5)

4. (18z2 + 2z + 6) – (–4z2 – 5z + 24)

5. (m + n + n2) – (8m2 – mn + n)

6. (6a + 7b2 + 8a3) – (–7a2 – 2a3 + 2b2)

7. (7 – 8e4 + 9e2) – (–e4 – e2 + 10)

8. (t – st + 3s) – (–st + t – s)

continued


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Solve problems 9–14 by setting up a subtraction of polynomials.

9. The area of a plot of land including a house can be represented by the polynomial 2x2 + 4x, and the area of just the yard can be represented by the polynomial 3x + 3. What is the area of the house, excluding the yard?

10. A projectile is shot from point P, a distance described by the polynomial 9t 2 + 8t + 1. The projectile is then shot from the point where it landed back to point P, but this time it only travels the distance described by the polynomial 3t 2 + 10t. How far from the initial point P is the projectile?

11. Julia owns a musical instrument store. She buys guitars for $350, saxophones for $420, and drum sets for $520. Then she sells the guitars for $499, the saxophones for $599, and the drum sets for $699. What is her profit if she buys and sells g guitars, s saxophones, and d drum sets?

12. Patrick is making tarts to sell. It costs him $8 to make each almond tart, $7.50 to make each pumpkin tart, $11.50 to make each key lime tart, and $13.50 to make each chocolate tart. He plans to sell the almond and pumpkin tarts for $12 each, and the key lime and chocolate tarts for $16 each. If he sells a almond tarts, p pumpkin tarts, k key lime tarts, and c chocolate tarts, what is his profit?

13. An unmanned test rocket is programmed to travel a distance of 8t 2 – 2t – 6. However, current strong winds will slow it down by a distance described with the polynomial 5t 2 – 6t + 1. If the rocket is launched with the current winds, how far will it travel?

14. A house with a yard, an outdoor shed, and a parking area will be built on a piece of land with an area of 80a2 + 25a + 100. The yard, the outdoor shed, and the parking area will occupy an area of 15a2 + 8a + 64. What is the area that the house itself will cover?


naMe:


Independent Practice Subtract the following polynomials.

1. (11a – 12b – 13) – (–17a + 19b – 21)

2. (–9p + 3q – 27r) – (–15p + 14q – 13r)

3. (–10g + 6h – 3gh) – (15g + 5h + 3gh)

4. (–2z2 + 5z + 7) – (z2 + 5z + 2)

5. (–2m + n2 + m2) – (–mn + n2 + m2)

6. (–ab – 3b + 5a) – (–7a – 2ab + b)

7. (k3 – k4 + 81k2) – (–24k4 + k3 – 1)

8. (pq – q + 3p) – (–p + pq – q)

continued


naMe:




9. The lot where a new building and its parking lot will be constructed has an area of 16x2 + 8x + 4. The area that the parking lot will occupy is 2x2 + 6x + 4. What will be the area of the building?

10. Edrick throws a football a distance described by the polynomial 12t 2 + 18t + 1. His little brother Manny catches it, but is too little to throw it all the way back to Edrick. The football only travels the distance described by the polynomial 4t 2 + 21t. How far from Edrick did the football land?

11. At his record store, Pedro sells vinyl records for $8.99 each, CDs for $11.99 each, and DVDs for $16.99 each. The wholesale price at which Pedro buys these products is $6 for vinyl records, $10.50 for CDs, and $15 for DVDs. If he buys and sells r records, c CDs, and d DVDs, what is his profit?

12. Shawn is making pies to sell. It costs him $12.50 to make each blueberry pie, $11 for each strawberry pie, $14.50 for each raspberry pie, and $7 for each apple pie. He plans to sell the apple pies for $12 each and all the other kinds of pies for $18 each. If he sells b blueberry pies, s strawberry pies, r raspberry pies, and a apple pies, what is his profit?

13. A weather balloon is programmed to travel a distance of 24t 2 – 24t + 10. However, it was filled with the wrong type of gasoline, which will slow it down by a distance described with the polynomial 5t 2 – 14t + 3. How far will the weather balloon travel?

14. A mini-golf course with a picnic area, a tool shed, and a parking area will be built on a piece of land with an area of 10a2 + 16a + 36. The picnic area, the tool shed, and the parking area will occupy an area of 2a2 + 3a + 4. What is the area that the mini-golf course itself will cover?


naMe:


Assessment

Progress AssessmentSubtract the following polynomials.

1. (3d – 8b + a – 4c) – (5a + 11b – 4c + 3d – 9)

2. (10y – y2 + 4y3 + 8) – (3 + 2y – y2 – 4y3)


3. Rafiq is selling berries in his fruit stand. He bought strawberries for $3.25 a pound, blueberries for $4.15 a pound, and raspberries for $6.45 a pound. He sells each pound of strawberries for $3.95, each pound of blueberries for $4.95, and each pound of raspberries for $6.95. What is his profit if he buys and sells s pounds of strawberries, b pounds of blueberries, and r pounds of raspberries?

4. Veronica is drawing the blueprint for a house. The area of the outdoor garden can be represented by 5p2 + 15p + 10. In the garden, there will be a circular pool with an area of 3p2, and a deck with an area of 6p. The rest of the garden will be covered with grass. How big is the area that will be covered with grass?

5. Ren and Martin throw one ball each up into the air. The distance traveled by Ren’s ball is 11d 2 – 5d + 8, and the distance traveled by Martin’s ball is 4d 2 + 12d + 1. What is the difference between the distances traveled by Ren’s ball and Martin’s ball?

6. A robot factory has w workers and has invested k dollars in machinery. The number of robots that the factory produces in a day is determined by the polynomial 10w + 100kw + k2. Of those robots, 15kw + 2w are defective. How many non-defective robots does the factory produce in a day?




Instruction

Resource List• Roberts, Donna, for Oswego City (NY) School District Regents Exam Prep Center.

“Subtracting Polynomials.”

www.regentsprep.org/regents/math/algebra/AV2/sp_subt.htm

This Web site has an instructive step-by-step explanation on how to subtract polynomials, using colors and algebra tiles to ensure student understanding.

• Russell, Deb, for About.com. “Addition and Subtraction of Polynomials.”

www.math.about.com/library/blpoly.htm

This Web site explains how to subtract polynomials step-by-step, from finding the additive inverse and rewriting the subtraction as a sum, to grouping like terms to add.

• SparkNotes. “Addition and Subtraction of Polynomials.”

www.sparknotes.com/math/algebra1/polynomials/section2.rhtml

This Web site has examples and exercises of both subtraction and addition of polynomials.



Lesson 4 Answer KeyLesson Pre-Assessment, p. 74 1. –2xy + 6y + 5 4. 0.433t 2 – 0.043t + 0.276 2. 2x2 + 14x + 12 5. 34x + 1,655 3. 2y2 + 4y

Guided Practice, p. 89 1. –96a – 4b – 32ab 8. 4s 2. 18p – 22q – 25r 9. 2x2 + x – 3 3. –6g – 10 10. 6t 2 – 2t + 1 4. 22z2 + 7z – 18 11. 179d + 149g + 179s 5. –8m2 + mn + m + n2 12. 4a + 2.5c + 4.5k + 4.5p 6. 10a3 + 7a2 + 6a + 5b2 13. 3t 2 + 4t – 7 7. –7e4 + 10e2 – 3 14. 65a2 + 17a + 36

Independent Practice, p. 91 1. 28a – 31b + 8 8. 4p 2. 6p – 11q – 14r 9. 14x2 + 2x 3. –25g – 6gh + h 10. 8t 2 – 3t + 1 4. –3z2 + 5 11. 1.49c + 1.99d + 2.99r 5. –2m + mn 12. 5a + 5.5b + 7s + 3.5r 6. 12a + ab – 4b 13. 19t 2 – 10t + 7 7. 23k 4 + 81k 2 + 1 14. 8a2 + 13a + 32

Progress Assessment, p. 93 1. –4a – 19b – 9 4. 2p2 + 9p + 10 2. 8y3 + 8y + 5 5. 7d 2 – 17d + 7 3. 0.80b + 0.70s + 0.50r 6. k 2 + 85kw + 8w

Unit 2 • operations with algebraic expressionsLesson 5: Multiplying Monomials

naMe:



Assessment

Lesson Pre-AssessmentSolve the following problems.

1. (4y2)(5y)

2. (3x)(7)(10x2)

3. (z3)(z)(4y2)(y)(2x2)

4. The area of a circle is πr2, where π = 3.14. What is the total area of 5 circles if each of them has the same radius, r ?

5. A living room has a length of 8x, a width of 5x, and a height of 6x. What is the volume of the living room?



Instruction

Lesson 5: Multiplying MonomialsEssential Questions 1. Is applying properties of exponents to algebraic expressions the same as applying them to

numbers?

2. How is multiplying monomials similar to multiplying numbers?

3. What real-world situations might require solving multiplications of monomials?

WORDS TO KNOW

base in the expression xn, x is the base and indicates the number that is going to be multiplied by itself n times


commutative property in a multiplication problem, the product remains the same even of multiplication if the order of the factors is changed







naMe:



Warm-Up Option 1Apply properties of exponents to simplify the following expressions. You do not need to solve.

1. 24 • 23

2. (62)5

3. (417 • 31.567)5

4. m12 • m6

5. (n9)2

6. (p • q)7


naMe:


Warm-Up Option 2Bebo picks a number and raises it to the fifth power, then multiplies the result by the same number raised to the third power. Tanya picks a different number and raises it to the fourth power, then squares the result. The two of them then multiply their results.

1. What is the expression that results when Bebo picks a number and raises it to the fifth power, then multiplies the result by the same number raised to the third power?

2. What is the expression that results when Tanya picks a number and raises it to the fourth power, then squares the result?

3. What happens when the two of them multiply their results?




Instruction

Warm-Up Option 1: Debrief 1. 24 • 23

• Remind students that 24 = 2 • 2 • 2 • 2 and 23 = 2 • 2 • 2.

• To find the product of 24 • 23, multiply 2 • 2 • 2 • 2 and 23 = 2 • 2 • 2.

• When bases are the same, the multiplication of powers can be rewritten as one power with the exponents added.

• The general rule to multiply powers of the same base is to add the exponents.

24 • 23 = 24 + 3

24 • 23 = 24 + 3 = 27

2. (62)5

• Remind students that 62 = 6 • 6.

• The problem can now be thought of as (6 • 6)5 or (6 • 6) • (6 • 6) • (6 • 6) • (6 • 6) • (6 • 6) or 610.

• The general rule to find the power of a power is to multiply the exponents.

• Rewrite the power of a power with the exponents multiplied.

(62)5 = 62 • 5 = 610

3. (417 • 31.567)5

• To solve a power of products, rewrite it as a product of powers.

• Rewrite the expression by “distributing” the exponent to the factors in the base of the power.

(417 • 31.567)5 = 4175 • 31.5675

• It is possible that students feel intimidated by exercises with variables, even though they are essentially the same as those with only numbers.

• Point out to students that the following exercises are just as easy as the previous ones.



Instruction

4. m12 • m6

• Just like with numerical expressions, to multiply powers of the same base of algebraic expressions, add the exponents.

• Rewrite the multiplication of powers as one power with the exponents added.

m12 • m6 = m12 + 6

m12 • m6 = m12 + 6 = m18

5. (n9)2

• To find the power of a power, multiply the exponents.

• Rewrite the power of a power with the exponents multiplied.

(n9)2 = n9 • 2

(n9)2 = n9 • 2 = n18

6. (p • q)7

• To solve a power of products, rewrite it as a product of powers.

• Rewrite the expression by “distributing” the exponent to the factors in the base of the power.

(p • q)7 = p7 • q7




Instruction

Warm-Up Option 2: Debrief 1. What is the expression that results when Bebo picks a number and raises it to the fifth power,

then multiplies the result by the same number raised to the third power?

• First, assign a variable to the number Bebo picks; for example, b.

• Raise the number to the fifth power: b5

• Multiply by b raised to the third power, or multiply by b3: b5 • b3

• Remind students that b5 = b • b • b • b • b and b3 = b • b • b.

b5 • b3 = b • b • b • b • b • b • b • b or b8

• The general rule to multiply powers of the same base is to add the exponents.

b5 • b3 = b5 + 3 = b5 + 3 = b8

The expression that results is b8.

2. What is the expression that results when Tanya picks a second number and raises it to the fourth power and then squares the result?

• Assign a variable to the number Tanya picks; for example, t.

• Raise the number to the fourth power: t 4

• Square the result: (t 4)2

• Remind students that t 4 = t • t • t • t.

• The problem can now be thought of as (t • t • t • t)2 or (t • t • t • t) • (t • t • t • t) or t 8.

• The general rule to find the power of a power is to multiply the exponents.

(t 4)2 = t 4 • 2

• Rewrite the power of a power with the exponents multiplied: t 8

The expression that results is t 8.



Instruction

3. What happens when the two of them multiply their results?

• Write an expression for the product of both results: b 8 • t 8

• You can further simplify the expression by rewriting the product of powers as a power of products.

b8 • t 8 = (bt) 8

After multiplying, the result is (bt) 8.

• A common mistake is that students try to add the exponents. This is not allowed because the bases are not the same.


naMe:



Focus ProblemAnn, a bookmaker, is recruiting volunteers to help her print books that she will distribute for free to students who cannot afford them. Each volunteer can print 3ab books per day, and she recruits b2c2 volunteers. She decides that she needs to keep this operation running for 5a2bc4 days to print all the books needed.

1. How many books can her entire team of volunteers print per day?

2. How many books will be printed in all?

3. If Ann switches to a much simpler book design that allows each volunteer to print 7a2b books per day, how many books can her team of volunteers print per day?

4. How many of these simpler books will the team print in all if Ann keeps the operation running for the same number of days as before?



Instruction


Present the problem to students and ask them which operation(s) are needed to find the number of books that Ann wants to print. (multiplication)

Point out to students that in the problem, three amounts are given in monomial form.

• First, point at the monomial 3ab and ask them what amount this expression represents. (the amount of books that each volunteer can print in one day)

• Then, point at the monomial b2c2 and ask what it represents. (the number of volunteers recruited)

• Finally, point at the monomial 5a2bc4 and ask what it represents. (the number of days the operation is running)

Tell students that in order to solve this problem they will have to multiply monomials.

• Guiding students with questions, explain that multiplying the number of books one volunteer can print in one day by the number of volunteers will yield the number of books printed per day by the whole team of volunteers.

• Guiding them with questions, explain that multiplying the number of books printed in one day by the number of days will give students the number of books printed in all.

Example

The Department of Transportation determines that the number of pedestrians crossing a bridge is 6p2 each day. How many pedestrians will cross the bridge over a period of 12p3 days?

Solution

• To find the total number of pedestrian crossing the bridge, multiply the number of pedestrians per day by the number of days: 6p2 • 12p3

• Explain to students that in order to multiply to monomials, they must first multiply the coefficients, then the variables.

• Remind them that by the commutative property of multiplication:

6p2 • 12p3 = 6 • 12 • p2 • p3




Instruction

• Multiply the coefficients.

6 • 12 • p2 • p3 = 72 • p2 • p3

• Remind students that in order to multiply two powers with the same base, add the exponents.

72 • p2 • p3 = 72 • p2 + 3 = 72p5

A total of 72p5 pedestrians will cross the bridge.


Ann, a bookmaker, is recruiting volunteers to help her print books that she will distribute for free to students who cannot afford them. Each volunteer can print 3ab books per day, and she recruits b2c2 volunteers. She decides that she must keep this operation running for 5a2bc4 days to print all the books needed.

Question 1

How many books can her entire team of volunteers print per day?

Instruction

• In order to find the answer, multiply the number of books each volunteer can print per day by the number of volunteers in the team.

• Elicit from students that each volunteer can print 3ab books per day.

• Elicit from students that there are b2c2 volunteers in the team.

• The number of books the team can print per day is: 3ab • b2c2

• Remind students that b = b1.

• Simplify the product of powers with the same base, b, by adding the exponents.

3ab • b2c2 = 3ab1 • b2c2 = 3ab1 + 2c2 = 3ab3c2

• Students having difficulty with this concept may need to see the problem written out:

(3ab) • (b2c2) = (3 • a • b) • (b • b • c • c)



Instruction

• Removing the parentheses results in 3 • a • b • b • b • c • c.

• Students can now count the occurrences of the variables to determine the exponents.

3 • a • b • b • b • c • c = 3ab3c2

The team of volunteers can print 3ab3c2 books per day.

Question 2

How many books will be printed in all?

Instruction

• To find the answer, multiply the number of books that can be printed per day (found in the previous question) by the number of days that the volunteers will work.

• The number of books that can be printed per day is 3ab3c2.

• The number of days that the volunteers will work is 5a2bc4.

• The number of books that Ann will print in all can be found by solving the following product:

3ab3c2 • 5a2bc4

• First, multiply the coefficients, then multiply powers with the same base.

3ab3c2 • 5a2bc4 = 15 • a • a2 • b3 • b • c2 • c4

• To multiply powers with the same base, add the exponents.

• Remind students that the exponent of any base that doesn’t have a power can be expressed as 1 (e.g., a = a1).

15 • a • a2 • b3 • b • c2 • c4 = 15 • a1 + 2 • b3 + 1 • c2 + 4

• Add the exponents.

15 • a1 + 2 • b3 + 1 • c2 + 4 = 15 • a3 • b4 • c6




Instruction

• Again, students having difficulty understanding this rule may want to write out each exponent.

• Group like terms as you expand the equation.

3ab3c2 • 5a2bc4

= (3 • a • b3 • c2) • (5 • a2 • b • c4)

= 3 • a • b3 • c2 • 5 • a2 • b • c4

= 3 • 5 • a • a2 • b3 • b • c2 • c4

= 15 • a • a2 • b3 • b • c2 • c4

= 15 • a • a • a • b • b • b • b • c • c • c • c • c • c

= 15 • a3 • b4 • c6

Ann will print 15a3b4c6 books in total.

Question 3

If Ann switches to a much simpler book design that allows each volunteer to print 7a2b books per day, how many books can her team of volunteers print per day?

Instruction

• In order to find the answer, multiply the number of books each volunteer can print per day by the number of volunteers in the team.

• The number of books the team can print per day is 7a2b • b2c2.

• Remind students that b = b1.

• Simplify the product of powers with the same base, b, by adding the exponents.

7a2b • b2c2 = 7a2b1 • b2c2 = 7a2b1 + 2c2 = 7a2b3c2

The team of volunteers can now print 7a2b3c2 books per day.

• Again, if students are having difficulty with this concept, encourage them to write out the exponents until they become more comfortable with the rule.




How many of these simpler books will the team print in all if Ann keeps the operation running for the same amount of days as before?

Instruction

• To find the answer, multiply the number of books that can be printed per day (found in the previous question) by the number of days that the volunteers will work.

• The number of books that can be printed per day is now 7a2b3c2.

• The number of days that the volunteers will work is the same, 5a2bc4.

• The number of books that will be printed in total can be found by solving the following product:

7a2b3c2 • 5a2bc4

• First, multiply the coefficients; then multiply powers with the same base.

7a2b3c2 • 5a2bc4 = 35 • a2 • a2 • b3 • b • c2 • c4

• To multiply powers with the same base, add the exponents.

35 • a2 • a2 • b3 • b • c2 • c4 = 35 • a2 + 2 • b3 + 1 • c2 + 4


35 • a2 + 2 • b3 + 1 • c2 + 4 = 35 • a4 • b4 • c6

Ann will print 35a4b4c6 books in all.




Instruction


Multiply the following monomials: 16x3y5z8 • 3z7x2

Solution

• To solve the problem, first multiply the coefficients.

16x3y5z8 • 3z7x2 = 16 • 3 • x3y5z8 • z7x2 = 48 • x3y5z8 • z7x2

• By the commutative property of multiplication, group powers with the same base.

48 • x3y5z8 • z7x2 = 48 • x3 • x2 • y5 • z8 • z7

• Then multiply powers with the same base by adding the exponents.

48 • x3 • x2 • y5 • z8 • z7 = 48 • x3 + 2 • y5 • z8 + 7


48 • x3 + 2 • y5 • z8 + 7 = 48 • x5 • y5 • z15

16x3y5z8 • 3z7x2 = 48x5y5z15

Example 2

Simplify the following expression: (2a4)2 • 3a4

Solution

• Simplify by applying properties of exponents to monomials, following the order of operations.

• Start by simplifying the first factor.

• A power of a product is a product of powers.

(2a4)2 = (2)2 • (a4)2 = 4 • (a4)2



Instruction

• Then, to solve the power of a power, multiply exponents.

4 • (a4)2 = 4 • a4 • 2 = 4a8

• The expression is now 4a8 • 3a4.

• Multiply coefficients and then the powers with the same base by adding exponents.

4a8 • 3a4 = 4 • 3 • a8 • a4 = 12 • a8 • a4 = 12a8 + 4 = 12a12

(2a4)2 • 3a4 = 12a12


naMe:



Guided PracticeMultiply the monomials.

1. a7b6c5 • d4c3b2

2. 18x3z • y4 • 3yz3x10

3. v5w6 • w3v • vw4 • v

4. 4h4 • 3j 3 • 2k 2 • hjk

Simplify the following expressions.

5. (a3b2c)3

6. (5x4)2 • (2x4)5

7. (w3)3 • (v2)2 • (wv2)5

8. (mn)3 • m6n • (n3)3

continued


naMe:


Use multiplication to solve the following problems.

9. A group of (5ab)2 students is organizing a party. Each of them will invite (a6b)3 guests. How many guests will they invite in total?

10. A tree house is (mn)4 long, (n3)5 wide, and m4n high. What is the volume of the tree house?

11. There are (4xy)3 tables at a restaurant. Each table has a basket with x2y slices of bread. How many slices of bread are there in total?

12. Jessica runs at a speed of 3xy2 meters per second for 9x2y seconds. What is the total distance she runs?

13. A skateboard factory produces 9ab4c2 skateboards and sells them for 10b3c3 each. What is the total revenue from those sales?

14. As a fundraiser, 8m6n2 volunteers sell an average of 2n4 boxes of cookies each. Each box of cookies is sold for 4m dollars. How much money was raised?


naMe:



Independent Practice Multiply the monomials.

1. w3x4y5 • z9y8x7

2. 7s4t 4 • 11t 5 • 2t 3s2

3. f 2h7 • h6g 3 • g 4f 5

4. mn2pq2 • 2q4m6 • 13pn4q5


5. (p5q4)6 • (p8q10)3

6. r 11s12t 13 • (5r 3)3 • (st)9

7. (x12y10)4 • (x5)5 • (y3)3

8. (ef )x • (f 2)x

continued


naMe:


Use multiplication to solve the following problems.

9. A ski factory produces kw2t 3 skis per month. The price of the skis is (4t 4w)2. If in December the factory sells all the skis they produce, what will be the revenue for skis that month?

10. Carla is riding her bike. She rides without stopping for r 2 hours at a constant speed of 5s4r miles per hour. How many miles does she ride?

11. A pool is 5p meters long, 6p meters wide, and 7q meters deep. What is the volume of the pool?

12. Matthew’s restaurant has served 6mn3 tables of customers today, and has served 8m2 bottles of water per table. How many total bottles of water has Matthew’s restaurant served today?

13. A high school has 11yz5 senior students. Each one brings 5xy guests to the graduation ceremony. How many guests in all do the students bring to the ceremony?

14. As a fundraiser, 6a5b2c volunteers are selling T-shirts. Each student sells 6bc3 T-shirts at a price of 3a2c2 each. How much do they raise in total?


naMe:



Assessment

Progress AssessmentSolve the following problems.

1. (a7b)3 • ac3 • (2b2)5

2. v2w2zy4 • x3v2y2 • z3x2 • vy10x8

3. A school district has h schools with s students each. Each year, they distribute h4s5 sheets of paper per student. How many sheets of paper does the school district distribute in a year?

4. A community garden 4a yards wide and 5b yards long produces (ab)3 pounds of food per square yard each season. How many pounds of food does the community garden produce each season?

5. Cassandra plays a nationwide tour with her jazz band. She travels through x states, playing x2 venues in each of those states, and sells (2xp)3 CDs in each venue. How many CDs does she sell?



Instruction

Resource List• edHelper.com. “Multiplying Exponents.”

www.edhelper.com/exponents5.htm

This sample worksheet with 26 monomial multiplication problems includes multiplication and properties of exponents applied to monomials.

• Keeler, Alice, for Quia. “Multiplying Monomials.”

www.quia.com/rr/79718.html

Use this activity to practice multiplying monomials. Three hints total are allowed for the game. Incorrect answers result in the end of the game.




Lesson 5 Answer KeyLesson Pre-Assessment, p. 96 1. 20y3 4. 15.7r 2. 210x3 5. 240x3

3. 8x2y3z4

Guided Practice, p. 112 1. a7b8c8d4 8. m9n13

2. 54x13y5z4 9. 25a20b5

3. v8w13 10. m8n20

4. 24h5j 4k3 11. 64x5y4

5. a9b6c3 12. 27x3y3

6. 800x28 13. 90ab7c5

7. w14v14 14. 64m7n6

Independent Practice, p. 114 1. w3x11y13z9 8. exf 3x

2. 154s6t 12 9. 16kw4t 11

3. f 7h13g7 10. 5s4r3

4. 26m7n6p2q11 11. 210p2q 5. p54q54 12. 48m3n3

6. 125r20s21t 22 13. 55xy2z5

7. x73y49 14. 108a7b3c6

Progress Assessment, p. 116 1. 32a22b13c3 4. 20a4b4

2. v5w2x13y16z4 5. 8x6p3

3. h5s6

Unit 2 • operations with algebraic expressionsLesson 6: Dividing Monomials

naMe:


Assessment

Lesson Pre-AssessmentSolve the following problems.

1. 9y3 ÷ 3y

2. 28x7 ÷ 7x2

3. 2z3 ÷ 8z4

4. A garden with an area of 12m2 is subdivided in plots, each with an area of 4m2. How many plots will the garden have?

5. Eddie makes 64a3 cups of cake batter. He is going to bake the cakes in baking pans that have a volume capacity of 16a3 cups. How many baking pans will he need?

Accuplacer College-Ready Mathematics: Elementary Algebra120



Instruction

Lesson 6: Dividing MonomialsEssential Questions 1. How does solving arithmetic division compare to algebraic division?

2. What real-world situations might require solving divisions of monomials?

WORDS TO KNOW






negative exponent an exponent with a negative sign in front of it; indicates how many times to divide by a number





naMe:


Warm-Up Option 1Apply properties of exponents to simplify the following expressions. You do not need to solve.

1. 16 • 2–3

2. 33

7

4

3. 162 • 3–4

4. 55

8

10

5. a6 • b–6

6. kk

13

5


naMe:



Warm-Up Option 2Roberta picks a number as the base of a power and raises it to the exponent –5, then multiplies it by the same number raised to the exponent 3. Keith picks a different number as the base of a power and raises it to the exponent 2. Then they take their results and multiply them.

1. What is the result when Roberta picks a number as the base of a power and raises it to the exponent –5, then multiplies it by the same number raised to the exponent 3?

2. What is the result when Keith picks a different number as the base of a power and raises it to the exponent 2?

3. What happens when Roberta and Keith multiply their results?



Instruction

Warm-Up Option 1: Debrief 1. 16 • 2–3

• A negative exponent indicates how many times to divide by a number. In this case, the exponent –3 indicates that 16 must be divided 3 times by 2.

• To solve, first rewrite the multiplication as a division and move the power with the negative exponent to the denominator, changing the sign of the exponent.

16 • 2–3 = 1623

• Now, simplify the power.

23 = 8, so 162

1683 =

16 • 2–3 = 168

2. 33

7

4

• To divide powers with the same base, subtract the exponents: 33

37

47 4= −

• Simplify the difference.

37 – 4 = 33

33

7

4 = 33

• If students are having difficulty remembering this rule, it is often helpful to expand the exponents and cancel appropriately.

33

3 3 3 3 3 3 33 3 3 3

3 3 3 3 3 3 33 3 3 3

7

4 =• • • • • •

• • •=

• • • • • •• • •

= 33 3 3 33• • =




Instruction

3. 162 • 3–4

• To simplify this expression, first rewrite it as a division by moving the power with the negative exponent to the denominator, changing the sign of the exponent.

162 • 3–4 = 16234

• Now, simplify the power.

34 = 81, so 1623

162814 =

162 • 3–4 = 16281

4. 55

8

10

• To divide powers with the same base, subtract the exponents: 55

58

108 10= −


58 – 10 = 5–2

• Since the exponent is negative, we need to rewrite the expression as a fraction.

5–2 = 152

• If students are having difficulty remembering this rule, it is often helpful to expand the exponents and cancel appropriately.

55

5 5 5 5 5 5 5 55 5 5 5 5 5 5 5 5 5

5 5 58

10 =• • • • • • •

• • • • • • • • •=

• • • 55 5 5 5 55 5 5 5 5 5 5 5 5 5

15 5

152

• • • •• • • • • • • • •

=•

=

55

15

8

10 2=



Instruction

5. a6 • b–6

• To simplify this expression, first rewrite it as a division by moving the power with the negative exponent to the denominator, changing the sign of the exponent.

a6 • b–6 = ab

6

6

• Since the exponents on numerator and denominator are the same, you can further simplify this expression using the property of power of a fraction.

ab

ab

6

6

6

=

• It is possible that students feel intimidated by exercises with variables, even though they are essentially the same as those with only numbers. Point out that these exercises follow the same rules as the previous exercises, but solving the exponentiation is not necessary.

6. kk

13

5

• To divide powers with the same base, subtract the exponents.

kk

k13

513 5= −


k13 – 5 = k8

• If students are having difficulty remembering this rule, expand the exponents and cancel appropriately.

kk

k k k k k k k k k k k k kk k k k k

k k k

13

5 =• • • • • • • • • • • •

• • • •

=• • • kk k k k k k k k k k

k k k k kk k k k k k k k

• • • • • • • • •• • • •

=• • • • • • •

=

1kk

k8

8

1=




Instruction

Warm-Up Option 2: Debrief 1. What is the result when Roberta picks a number as the base of a power and raises it to the

exponent –5, then multiplies it by the same number raised to the exponent 3?

• First, assign a variable to the number Roberta picks; for example, r.

• Roberta picks a number and raises it to the exponent –5: r –5

• Then, she multiplies by the same number raised to the exponent 3, or multiplies by r 3.

• The expression formed as a result is r –5 • r 3.

• A power with a negative exponent can be expressed as a fraction, moving the power to the denominator and changing the sign of the exponent.

r rrr

rr

− −• = = =5 33

52

2

1

• For students having difficulty remembering this rule, expand the exponents and cancel appropriately.

r rrr

r r rr r r r r

r r rr r r r r r r

− • = =• •

• • • •=

• •• • • •

=•

=5 33

5

1 112r

2. What is the result when Keith picks another number as the base of a power and raises it to the exponent 2?

• Assign a variable to the number Keith picks; for example, k.

• Keith picks a number and raises it to the exponent 2: k2

3. What happens when Roberta and Keith multiply their results?

• Set up an expression multiplying Roberta’s and Keith’s results: r –2 • k2

• Remember that when multiplying by a negative exponent, the exponent is telling us how

many times to divide by that base. Rewrite the product as division and change the sign of the

negative exponent: r kkr

− • =2 22

2

• Finally, you can express the division of powers as the power of a division.

kr

kr

2

2

2

=


naMe:


Focus ProblemThe length of a rectangular farm can be represented by the expression 3ab3c2, and its area is 18a5b2c4.

1. What is the width of the farm?

2. If the farm is divided into a2b2 parcels along its width, how wide is each parcel?

3. If the farm is divided into a2b2c2 parcels along its length, how wide is each parcel?

4. If the farm is divided into parcels that each have an area of 2a3b3c3, how many parcels is the farm divided into?




Instruction


Dividing monomials is similar to arithmetic division. In both situations, common factors of the numerator and denominator are canceled.

It is helpful to remind students of the process of arithmetic division.

• To divide 15 by 9, list the factors of each number.

159

5 33 35 33 353

=••

=••

=

• To divide the two monomials, first divide the coefficients as if the division were arithmetic.

• Then, divide powers with the same base by subtracting the exponents.

Use the following example to help demonstrate this concept.

Example

Divide the following monomials: 16m5n5 ÷ 23m2n7

Solution

• First, set up the division as a fraction: 162

5 5

3 2 7

m nm n

• Expand 23.

162

168

5 5

3 2 7

5 5

2 7

m nm n

m nm n

=

• Divide the coefficients 16 and 8.

168

25 5

2 7

5 5

2 7

m nm n

m nm n

=



Instruction

• Divide powers with the same base by subtracting the exponents.

22

5 5

2 75 2 5 7m n

m nm n= − − = 2m3n –2

• The power with a negative exponent means that we need to divide by the base the number of times indicated by that exponent. In other words, we can rewrite the expression as a fraction to avoid having a negative exponent.

223 2

3

2m nmn

− =

• Students often have difficulty understanding why this rule works.

• Expand the exponents and cancel common factors to illustrate the rule.

2 25 5

2 7

m nm n

m m m m m n n n n nm m n n n n n n

=• • • • • • • • • •

• • • • • • • • nnm m m m m n n n n nm m n n n n n n nm m

=• • • • • • • • • •

• • • • • • • •

=• •

2

2 •••

=

mn n

mn2 3

2

So, 16m5n5 ÷ 23m2n7 = 2 3

2

mn

.


The length of a rectangular farm can be represented by the expression 3ab3c2, and its area is 18a5b2c4.

Question 1

What is the width of the farm?

Instruction

• Remind students that the formula for the area of a rectangle is A = length • width = l • w.




Instruction

• To find the width of the farm, divide the area by the length.

A = l • w

wAla b cab c

=

=183

5 2 4

3 2

• To perform the division of monomials, first divide the coefficients.

• Then divide the powers with the same base by subtracting the exponents.

183

66 6

5 2 4

3 2

5 2 4

1 3 25 1 2 3 4 2 4a b c

ab ca b ca b c

a b c a b= = =− − − −−1 2c

• Expand the exponents and cancel common factors as in the example if students continue to misunderstand the rule for division.

The width of the farm is 6 4 1 2a b c− or 6 4 2a cb

.

Question 2

If the farm is divided into a2b2 parcels along its width, how wide is each parcel?

Instruction

• Divide the width of the farm by the number of parcels to find the width of each parcel.

6 4 1 2

2 2

a b ca b

−

• To perform the division of monomials, divide the powers with the same base by subtracting the exponents.

66 6

4 1 2

2 24 2 1 2 2 2 3 2a b c

a ba b c a b c

−− − − −= =



Instruction

• Remind students that the negative exponent means division.

662 3 2

2 2

3a b ca cb

− =

The width of each parcel is 6a2b–3c2, or 6 2 2

3

a cb

.

Question 3

If the farm is divided into a2b2c2 parcels along its length, how wide is each parcel?

Instruction

• Divide the length of the farm by the number of parcels to find the width of each parcel.

3 3 2

2 2 2

ab ca b c

• To perform the division of monomials, divide the powers with the same base by subtracting the exponents.

33 3

3 2

2 2 21 2 3 2 2 2 1 1 0ab c

a b ca b c a b c= =− − − −


• Also remind students that any number raised to the power of 0 equals 1.

331 1 0a b cba

− =

The width of each parcel is 3 1 1a b− or 3ba

.





If the farm is divided into parcels with an area of 2a3b3c3, how many parcels is the farm divided into?

Instruction

• Divide the area of the farm by the area of the parcels to find the number of parcels.

182

5 2 4

3 3 3

a b ca b c

• To perform the division of monomials, divide the coefficients first, and then divide the powers with the same base by subtracting their exponents.

182

9 95 2 4

3 3 35 3 2 3 4 3 2 1a b c

a b ca b c a b c= =− − − −


992 1

2

a b ca cb

− =

The farm is divided into 9a2b–1c or 9 2a cb

parcels.


Divide the following monomials: 27x5y–3z8 ÷ 6z3x9

Solution

• To solve, first divide the coefficients.

27

6

9

2

5 3 8

3 9

5 3 8

3 9

x y zz x

x y zz x

− −

=

• Then divide the powers with the same base by subtracting exponents.

92

92

5 9 3 8 3 4 3 5x y z x y z− − − − −=



Instruction

• The negative exponents indicate division: 92

92

4 3 55

4 3x y zzx y

− − =

27x5y–3z8 ÷ 6z3x9 = 92

5

4 3

zx y

Example 2

Divide the following monomials: 108m2p6q5 ÷ 12m–4n–4p4q4

Solution

• First, rewrite the division as a fraction.

108m2p6q5 ÷ 12m–4q4n–4p4 = 108

12

2 6 5

4 4 4 4

m p qm n p q− −

• Note to students that this time, there are negative exponents in the denominator.

• Explain to them that the exponents can be made positive by moving the power from the denominator to the numerator, or they can simply proceed to divide by subtracting exponents.

• Begin by dividing the coefficients.

108

12

92 6 5

4 4 4 4

2 6 5

4 4 4 4

m p qm n p q

m p qm n p q− − − −=

• Move the variables with negative exponents to the numerator and change the exponent to a positive number.

9 92 6 5

4 4 4 4

2 4 4 6 5

4 4

m p qm n p q

m m n p qp q− − =

• Simplify the numerator.

9 92 4 4 6 5

4 4

6 4 6 5

4 4

m m n p qp q

m n p qp q

=




Instruction

• Divide the powers with bases p and q.

9m6n4p6 – 4q5 – 4 = 9m6n4p2q1 = 9m6n4p2q

108m2p6q5 ÷ 12m–4n–4p4q4 = 9m6n4p2q

• The alternative method is to begin by dividing the coefficients.

108

12

92 6 5

4 4 4 4

2 6 5

4 4 4 4

m p qm n p q

m p qm n p q− − − −=

• Divide the powers with bases p and q by subtracting the exponents.

99 9

2 6 5

4 4 4 46 4 5 4

2

4 42

2

4

m p qm n p q

p qm

m np q

mm n− −

− −− − − −= = 44

• Then divide the power with the base m as usual, by subtracting exponents.

9 91

91

922

4 42 2 4

42 2 4

42p q

mm n

p qmn

p qmn

p qm− −− −

−+

−= = =( ) 664

1n−

• Remind students that when they have a negative exponent in the numerator, they move the power to the denominator.

• Similarly, when there is a negative exponent in the denominator, they can simply move the power to the numerator.

91

91

92 64

2 64

2 6 4p qmn

p qmn

p qm n− = =

108m2p6q5 ÷ 12m–4n–4p4q4 = 9m6n4p2q


naMe:


Guided PracticeDivide the following monomials.

1. (28a2b6c7) ÷ (84a5b2d 2)

2. (m5np2q3) ÷ (m3n4p2q)

3. (7x7y7z7) ÷ (x7y3z10w7)

4. (g3h4j 5k6) ÷ (6h2g 6k3j 8)

5. x –2y10z4 ÷ x2y–2z5

6. q5r 3s–4 ÷ q2s8t –6

7. a2b–3c4d–5 ÷ d–3c –2b3a2

8. m–10n–10p–10 ÷ m–15n–10p–5

continued


naMe:



Solve the following word problems using division of monomials.

9. A pizzeria has 24f 4g3h10 pounds of pizza dough for the day. Each pizza requires 8f 2g4 pounds of dough. How many pizzas can the restaurant make with the dough it already has?

10. Last month, Quince spent 20a2b8 hours learning new songs on the piano. On average, each song took him 5b4c2 hours to learn. How many songs did he learn?

11. A pool has an area of 8x3y4z5 and a volume of 56x2y6z9. What is the depth of the pool?

12. Adrianna makes 135m4n9 dollars from selling 9m5n5 backpacks at the same price. For how much did she sell each backpack?

13. A video store has 144p2q3 DVDs distributed equally among 6p3q3 shelves. How many DVDs are there on each shelf?

14. Each month, a mail carrier delivers an average of 225a4b3c3 pounds of mail to an average of 25a6b4c houses. How many pounds of mail does each house receive on average?


naMe:


Independent Practice Divide the following monomials.

1. (108x16y18z20) ÷ (18w20x15y15z10)

2. (t 4s14r 3q13) ÷ (r9s8q7t 6)

3. (m16n8p4) ÷ (p10m12n8)

4. (a5b6c2d) ÷ (c10a3bd 2)

5. 3g –4h–2j 7 ÷ 51g3h–6j 2

6. 8a–2b–3c–4 ÷ a4b6c–8

7. m5n–7p–9q11 ÷ m3p–7q9

8. z–3y5x–2w5v ÷ xy2z–5v3

continued


naMe:



Solve the following word problems using division of monomials.

9. Each month, a dog kennel serves an average of 30p2e3b5 pounds of dog food to an average of 15e6b2p dogs. How many pounds of food does each dog eat on average each month?

10. A taxi company’s cabs use 1000a4b10c6 gallons of gasoline each week. There are 100a3b8c2 cabs. How many gallons of gas did each cab use on average over the course of one week?

11. A pottery maker has 148f 7g4h2 pounds of clay. Each piece of pottery requires 4f 3g5 pounds of clay. How many pieces of pottery can the pottery maker produce with the clay she already has?

12. Last month, Lesley spent 17a4b4 hours memorizing 2a4b3 haikus. On average, how long did it take her to memorize each haiku?

13. A rectangular golf course has an area of 30x2y2z and a length of 4x4yz. What is the width of the golf course?

14. Aaron has spent a total of 256m7n10 dollars on his record collection. If he has 64m10n7 records in his collection, how much did he pay for each record on average?


naMe:


Assessment

Progress AssessmentSolve the following problems.

1. ca–3b2d 7 ÷ a2b7c –4d –6

2. v4w–1x5z–1 ÷ z–1y–3xw2v4

3. A computer factory has made 56g18f 3 dollars in revenue from selling computers at a price of 7g10f 10 dollars each. How many computers did the factory sell?

4. Last year, the United States produced 10,000m15n5p10 pounds of waste just from discarded cellular phones. If each cellular phone weighs 10m2n3p4 pounds, how many cellular phones were discarded?

5. A dairy farm produces 5k9h6g3 liters of milk each month, and sells them for a total of g5h6k5 dollars. How much does each liter of milk cost?

6. The total payroll of a company was 35x3y5z7 for 14x4y4z4 employees. What was each employee paid if each one earned the same amount?




Instruction

Resource List• Education.com. “Help with: Division of Monomials.”

www.education.com/reference/article/pre-algebra-help-division-monomials/

This Web site provides a basic tutorial on how to divide monomials. It demonstrates with an example why we subtract exponents to divide powers with the same base, then offers a few examples and five exercises with an answer key.

• Phillips, Nancy, for Quia. “Laws of Exponents: Dividing Monomials.”

www.quia.com/jg/839435.html

Players can choose between a game of Concentration or a game of Matching to practice dividing monomials.

• Russell, Deb, for About.com. “Basic Algebra: Dividing Monomials.”

www.math.about.com/od/polynomials/ss/divmon.htm

This Web site connects division of monomials (algebra) with division of numbers (arithmetic) with a few simple examples.



Lesson 6 Answer KeyLesson Pre-Assessment, p. 119 1. 3y2 4. 3 2. 4x5 5. 4

3. 14z

Guided Practice, p. 135

1. b ca d

4 7

3 23 8.

mp

5

5

2. m qn

2 2

3 9. 3 2 10f h

g

3. 7 4

3 7

yz w

10. 4 2 4

2

a bc

4. h kg j

2 3

3 36 11.

7 2 4y zx

5. yx z

12

4 12. 15 4nm

6. q r ts

3 3 6

12 13. 24p

7. cb d

6

6 2 14. 9 2

2

ca b

Independent Practice, p. 137

1. 6 3 10

20

xy zw

8. w y zv x

5 3 2

2 3

2. q sr t

6 6

6 2 9. 2 3

3

pbe

3. mp

4

6 10. 10ab2c4

4. a bc d

2 5

8 11. 37 4 2f h

g

5. h jg

4 5

717 12.

172

b

6. 8 4

6 9

ca b

13. 152 2

yx

7. m qn p

2 2

7 2 14. 4 3

3

nm

Progress Assessment, p. 139

1. c da b

5 13

5 5 4. 1,000m13n2p6

2. x yw

4 3

3 5. 5 4

2

kg

3. 8 8

7

gf

6. 52

3yzx

Unit 2 • operations with algebraic expressionsLesson 7: Multiplying Polynomials

naMe:



Assessment

Lesson Pre-AssessmentMultiply the following algebraic expressions.

1. (a + 3b)(6a – 2b)

2. (12 – x2)(x2 + 3x – 5)

3. (y – 3)(y2 – 6y + 9)

4. Over the course of one month, Michelle goes swimming (x + 8) times and each time she swims (4x + 7) laps. How many laps does she swim in one month?

5. A library has its books divided in (a2 – 4ab – 6a – 10b + b2) sections, and each section has (5a + 8b) books. How many books does the library have in total?



Instruction

Lesson 7: Multiplying PolynomialsEssential Questions 1. How is multiplying polynomials different from multiplying monomials?

2. What is the distributive property, and how does it apply to polynomial multiplication?

3. What real-world situations can be represented with polynomial multiplication?

WORDS TO KNOW


distributive property when multiplying polynomials, multiply each term of the first factor by each term in the second factor: (a + b)(c + d) = ac + ad + bc + bd






naMe:



Warm-Up Option 1Apply the distributive property to the following problems.

1. 8(m + n)

2. (x – y) • k

3. c3 • 4c2

4. 5x2(6x + 12)

5. –8y3(10 + 4y – y2)


naMe:


Warm-Up Option 2Solve the following word problems using multiplication of polynomials.

1. 4 times the sum of a number and 8 equals 38. What is the number?

2. A pool has a length of x yards. It is 3 yards wider than it is long. What is the area of the pool in square yards?




Instruction

Warm-Up Option 1: Debrief 1. 8(m + n)

• To multiply the expression above, use the distributive property of multiplication, which states that for any real numbers a, b, and c, a(b + c) = ab + ac and a(b – c) = ab – ac.

• The number 8 outside the parentheses is multiplying both terms inside the parentheses.

• When you perform the multiplication, you must distribute the 8 to each term in the sum.

8(m + n) = 8 • m + 8 • n = 8m + 8n

• Students commonly multiply just the first term of the sum by 8, resulting in an incorrect answer of 8m + n. Help students by demonstrating that 8(m + n) states there are 8 groups of (m + n).

• To further illustrate this statement, write out 8 groups of (m + n):

(m + n) + (m + n) + (m + n) + (m + n) + (m + n) + (m + n) + (m + n) + (m + n)

• Students should now see 8m + 8n.

• It is also helpful for students to draw arrows indicating the multiplication that needs to happen.

8(m + n) = 8 • m + 8 • n = 8m + 8n

8(m + n) = 8m + 8n

2. (x – y) • k

• To multiply the difference (x – y) by the variable k, use the distributive property.

• Distribute k so that it multiplies each term in the difference (x – y).

(x – y) • k = x • k – y • k = xk – yk

(x – y) • k = xk – yk



Instruction

3. c3 • 4c2

• To multiply the two monomials above, multiply the coefficients first and then multiply powers with the same base by adding their exponents.

c3 • 4c2 = 4 • c3 • c2 = 4 • c3 + 2 = 4 • c5

c3 • 4c 2 = 4 • c 5 = 4c 5

4. 5x2(6x + 12)

• To multiply the expressions above, distribute the monomial 5x2 to multiply both terms in the sum.

• To multiply two terms with common variables, multiply coefficients first and then multiply powers with the same base by adding their exponents.

5x2(6x + 12) = 5x2 • 6x + 5x2 • 12 = 30x2 + 1 + 60x2 = 30x3 + 60x2

5x2(6x + 12) = 30x3 + 60x2

5. –8y3(10 + 4y – y2)

• To multiply the expressions above, distribute the monomial –8y3 to all terms inside the parentheses.

• To multiply two terms with common variables, multiply coefficients first and then multiply powers with the same base by adding their exponents.

• Remind students that when there are negative signs in the expressions being multiplied, take special care to write the correct signs when solving.

• It is possible that students will forget the negative sign. Remind them that the third term of the difference in the parentheses is –y2, and that –8y3 is multiplying not only y2 but also its sign.

–8y3(10 + 4y – y2) = (–8y3)(10) + (–8y3)(4y) + (–8y3)( –y2) = –80y3 – 32y4 + 8y5

–8y3(10 + 4y – y2) = –80y3 – 32y4 + 8y5




Instruction

Warm-Up Option 2: Debrief 1. 4 times the sum of a number and 8 equals 38. What is the number?

• Represent algebraically the expression “4 times the sum of a number and 8,” then simplify it.

• Assign the variable x to “a number.”

• Translate “4 times the sum of x and 8.”

• “Sum” indicates addition.

• “The sum of x and 8” indicates the use of parentheses.

4(x + 8)

• To multiply, distribute the number 4 to both terms in the sum.

4(x + 8) = 4 • x + 4 • 8 = 4x + 32

• Students commonly multiply just the first term of the sum by 4, resulting in an incorrect answer of 4x + 8. Help students by demonstrating that 4(x + 8) indicates there are 4 groups of (x + 8).

• To further illustrate this statement, write out 4 groups of (x + 8):

(x + 8) + (x + 8) + (x + 8) + (x + 8)

• Students should now see 4x + 32.

• It is also helpful for students to draw arrows indicating the multiplication that needs to happen.

4(x + 8) = 4 • x + 4 • 8 = 4x + 32

• Solve the equation 4x + 32 = 38 by isolating x.

4x + 32 = 38

4x = 6

x = 64 = 32

Four times the sum 32 and 8 is equal to 38.



Instruction

2. A pool has a length of x yards. It is 3 yards wider than it is long. What is the area of the pool in square yards?

• The length of the pool is x. The width of the pool is (x + 3). The area of the pool can be determined by multiplying those two dimensions.

Area = length • width

Area = x(x + 3)

• To determine the area, use the distributive property.

x (x + 3) = x • x + x • 3 = x2 + 3x

The area of the pool is x2 + 3x square yards.


naMe:



Focus ProblemAn architect is planning the design of a new home. She wants to build a house with the option of having a driveway on one side, and the option of having a backyard that would extend behind both the house and the driveway. She has three different plans.

1. The first plan is to make the house w yards wide and a driveway 5 yards wide next to the house. Both the house and the driveway will be l yards long. What would be the area of the house plus the driveway?

2. The second plan is to add a backyard behind the house and driveway. The backyard would be 16 yards long, and the house and driveway would have the same dimensions as in the first plan. What would be the area of the whole house, including the driveway and backyard?

3. The third plan is to make the base of the house a square with sides s, to keep the driveway 5 yards wide, and to make the backyard only 12 yards long. What would be the area of the whole house?

4. Eventually, the architect decides that the house will not have a driveway, and that it will take up two lots in order to be more spacious. The house will be w yards wide plus the 25 yards of the additional lot. It will be l yards long. The backyard will be as wide as the house and 18 yards long. What would be the area of the whole house, including the backyard?



Instruction


The steps to solve each question of the focus problem are similar and involve multiplying polynomials. Multiplying polynomials is very much like multiplying a polynomial by a monomial as seen in the Warm-Ups. Use the distributive property of multiplication, which states that when multiplying polynomials, you must multiply each term of the first factor by each term in the second factor: (a + b)(c + d) = ac + ad + bc + bd.

Example

Multiply: (3x + 4y)(16x + y)

• Each term of the first polynomial, (3x + 4y), is multiplied by each term of the second polynomial, (16x + y).

• One way for students to organize their thinking is to use the mnemonic device FOIL: First terms, Outside terms, Inside terms, and Last terms.

First terms: 3x • 16x = 48x2

Outside terms: 3x • y = 3xy

Inside terms: 4y • 16x = 64xy

Last terms: 4y • y = 4y2

• To organize the process, use arrows to indicate the pairs being multiplied.

• (3x + 4y)(16x + y) = 3x • 16x + 3x • y + 4y • 16x + 4y • y = 48x2 + 3xy + 64xy + 4y2

(3x + 4y)(16x + y) = 48x2 + 3xy + 64xy + 4y

• Some students may fail to realize that the expression 48x2 + 3xy + 64xy + 4y still requires further simplification.

• Prompt students and guide them to realize that 3xy and 64xy are like terms—terms with the same variable base raised to the same power—and must be added.

(3x + 4y)(16x + y) = 48x2 + 67xy + 4y2

The product of (3x + 4y)(16x + y) is 48x2 + 67xy + 4y2.




Instruction


An architect is planning the design of a new home. She wants to build a house with the option of having a driveway on one side, and the option of having a backyard that would extend behind both the house and the driveway. She has three different plans.

Question 1

The first plan is to make the house w yards wide and a driveway 5 yards wide next to the house. Both the house and the driveway will be l yards long. What would be the area of the whole house plus the driveway?

Instruction

For many students, a picture of the situation is helpful when determining area.

l

w 5

• Write an expression for both the length and the width of the whole house including the driveway.

Total length: l

Total width: w + 5

• Then use the formula Area = length • width to determine the area of the house including the driveway.

Area = (l)(w + 5)

• Use the distributive property to simplify the expression.

(l)(w + 5) = l • w + l • 5 = lw + 5l

The area of the house and driveway is lw + 5l.




The second plan is to add a backyard behind the house and driveway. The backyard would be 16 yards long, and the house and driveway would have the same dimensions as in the first plan. What would be the area of the whole house, including the driveway and backyard?

Instruction

• Begin with a drawing.

16

l

w 5

• Write an expression for both the length and the width of the house, driveway, and backyard.

Total length: l + 16

Total width: w + 5

• Then use the formula Area = length • width to determine the area of the house, driveway, and backyard.

Area = (l + 16)(w + 5)


• Use arrows to demonstrate the terms being multiplied.

(w + 5)(l + 16) = w • l + w • 16 + 5 • l + 5 • 16




Instruction

• Simplify the expression.

w • l + w • 16 + 5 • l + 5 • 16 = lw + 16w + 5l + 80

• It is common to write the terms of an expression in alphabetical order by the variables.

• The area of the house is lw + 5l + 16w + 80.

Question 3

The third plan is to make the base of the house a square with sides s, to keep the driveway 5 yards wide, and to make the backyard only 12 yards long. What would be the area of the whole house?

Instruction


12

s

s 5


Total length: s + 12

Total width: s + 5

• Then use the formula Area = length • width to determine the area of the house, driveway, and backyard.

Area = (s + 12)(s + 5)




Instruction


(s + 5)(s + 12) = s • s + s • 12 + 5 • s + 5 • 12


s • s + s • 12 + 5 • s + 5 • 12 = s2 + 12s + 5s + 60 = s2 + 17s + 60

The area of the house is s2 + 17s + 60.

Question 4

Eventually, the architect decides that the house will not have a driveway, and that it will take up two lots in order to be more spacious. The house will be w yards wide plus the 25 yards of the additional lot. It will be l yards long. The backyard will be as wide as the house and 18 yards long. What would be the area of the whole house, including the backyard?

Instruction


18

l

w 25


Total length: l + 18

Total width: w + 25

• Then use the formula Area = length • width to determine the area of house, driveway, and backyard.

Area = (w + 25)(l + 18)




Instruction



(w + 25)(l + 18) = w • l + w • 18 + 25 • l + 25 • 18


w • l + w • 18 + 25 • l + 25 • 18 = lw + 18w + 25l + 450

The area of the house is lw + 18w + 25l + 450.


Multiply: (x3 + x2)(x + 1)

Solution

• This example will allow students to practice applying FOIL again.

(x3 + x2)(x + 1) = x3 • x + x3 • 1 + x2 • x + x2 • 1


x3 • x + x3 • 1 + x2 • x + x2 • 1 = x4 + x3 + x3 + x2

• Again, point out for students that x3 and x3 are like terms and must be added.

x4 + x3 + x3 + x2 = x4 + 2x3 + x2

The product of (x3 + x2)(x + 1) is x4 + 2x3 + x2.




Multiply: (x + 2)(x2 + 5)

Solution

• To solve, multiply the two algebraic expressions using the distributive property and FOIL.

(x + 2)(x2 + 5)

= (x • x2) + (x • 5) + (2 • x2) + (2 • 5)

= x3 + 2x2 + 5x + 10

(x + 2)(x2 + 5) = x3 + 2x2 + 5x + 10

Example 3

Multiply: (a + b + 4)(x2 – 2)

Solution

• To solve, multiply the two algebraic expressions using the distributive property.

• Note that in this case, one of the factors has three terms.

• The process is similar; each term in the first polynomial, (a + b + 4), gets multiplied by each term of the second polynomial (x2 – 2).

• Remind students to take into account the effect of the minus sign on 2, which becomes –2.

(a + b + 4)(x2 – 2)

= (a)(x2) + (a)(–2) + (b)(x2) + (b)(–2) + (4)(x2) + (4)(–2)

= ax2 + bx2 + 4x2 – 2a – 2b – 8

(a + b + 4)(x2 – 2) = ax2 + bx2 + 4x2 – 2a – 2b – 8


naMe:



Guided PracticeMultiply the following polynomials.

1. (a + 5b)(5a + b) 6. (3x + 10)(3x – 10)

2. (10m – 12n)2 7. (a2 – b)(2a + 2b)

3. (2p + 3q)3 8. (9 – t 2)(9 + t 2)

4. (4x3 + 3x2 + 7x + 14)(x + 5) 9. (a2 + 1)(a + b)(a – b)

5. (9y2 + 13xy + 17x2)(x + y) 10. (a – b)(a2 – 2ab + b2)

continued


naMe:


Solve the following problems by multiplying polynomials.

11. Kyla is riding her bicycle across the whole country. She rides an average of 5 hours each day at a speed of (y2 + 7) miles per hour. After (y + 4) days, how much distance has she covered?

12. Dr. Paxton is checking a patient’s heart rate. She counts heartbeats for (x2 – m) minutes, after which she concludes that the patient has a rate of (x2 + m) heartbeats per minute. What is the total number of heartbeats she counted?

13. Ignacio is building a tree house in the shape of a cube. Each side of the tree house measures (3t + 1) feet. What is the volume of the tree house?

14. At a school, the number of students in each classroom is (g + b), and there are a total of (gb + 5) classrooms. What is the total number of students in the school?

15. A library has (f + n + t) books in each shelf, and has a total of a2 shelves. How many books does the library have?

16. Damian runs for (x – 9) minutes every day at a speed of (x + 20) yards per minute. How many yards does he run every day?


naMe:



Independent Practice Multiply the following polynomials.

1. (10 – 8b)(10 + 8b) 6. (4a2 + 9)(2a – 3)(2a + 3)

2. (4m2 – 24m + 36)(2m – 6) 7. (3y – z)(10 + z2 + y2)

3. (g + 4)3 8. (2v2 + 5w2)(3vw + 3)

4. (2 + x3)(3x3 + 4x2 + 5x + 6) 9. (a2 – 4a + 4)(a – 2)

5. (2x + 2y)(4x2 + 8xy + 4y2) 10. (x4 + y6)(x2 + y3)(x2 – y3)

continued


naMe:


Solve the following problems by multiplying polynomials.

11. Kate is writing a novel. She is typing an average of (4k2 – 36k + 81) words each day, and has been working on it for (2k – 9) days now. How many pages has she written?

12. Emilio is studying for an exam. For (x – 3) days, he has been solving (x + 17) problems per day. How many problems has he solved so far?

13. Mia is organizing a fund-raiser raffle. She has (y – 9) volunteers selling (y – 11) raffle tickets each. If all the volunteers sell all the tickets they have, how many raffle tickets will have been sold?

14. At a state park, the number of picnic tables being used is (p3 + 14). Each of those picnic tables is being used by (p3 – 14) people. How many people are using picnic tables in total?

15. The length of an empty lot is (a2 + 16). Its width is (a2 – 16). What is the area of the lot?

16. Ruben rides his skateboard for (x – 12) minutes at an average speed of (x + 24) meters per minute. What is the distance that he rides?


naMe:



Assessment

Progress AssessmentMultiply the following algebraic expressions.

1. (5x – y)(25x2 – 10xy + y2)

2. (25 + a2)(5 + a)(5 – a)

3. Rachel is on a road trip. She has driven (7x + 2) miles. Her car consumes (x + 3) gallons of gas per mile. Each gallon of gas costs (x + 1) dollars. How many dollars has she spent on her road trip?

4. Rosa wants to build an outdoor sauna in her backyard. She wants it to be in the shape of a cube, with sides (4a – 5) feet long. What will the volume of the sauna be?

5. Maurice, an astronomer, has identified (y – 13) stars in each of (y + 4) galaxies. How many stars has he identified in total?

6. The space assigned to each vendor in the farmers’ market has width of x on the front and also a length of x from the front to the back. In addition, each vendor has 10 feet behind their assigned space to store boxes and other materials, and each vendor also has an additional width of 3 feet on one side of their stand, to make room between them and the next stand. What is the total area assigned to each vendor?



Instruction

Resource List• Coolmath.com. “Algebra Crunchers—Polynomials.”

www.coolmath.com/algebra/algebra-practice-polynomials.html

Choose from several sets of practice problems. The user is given an expression and is asked to work out the problem on paper. Click the “What’s the answer?” button to check results.

• Discovery Education: WebMath.com. “Multiply Polynomials.”

www.webmath.com/polymult.html

This Web site allows students to enter their own multiplication of polynomials problem and gives them a step-by-step answer. The answers provided can sometimes be hard to read, but nonetheless constitute a useful learning resource that students can use to check their work.

• Stapel, Elizabeth, for Purplemath. “Simple Polynomial Multiplication.”

www.purplemath.com/modules/polymult.htm

This Web site provides a complete tutorial on multiplication of polynomials, from multiplying monomials, to multiplying a monomial and a polynomial, and finally multiplying polynomials.




Lesson 7 Answer KeyLesson Pre-Assessment, p. 142 1. 6a2 + 16ab – 6b2

2. –x4 – 3x3 + 17x2 + 36x – 60 3. y3 – 9y2 + 27y – 27 4. 4x2 + 39x + 56 5. 5a3 – 12a2b – 27ab2 + 8b3 – 30a2 – 80b2 – 98ab

Guided Practice, p. 158 1. 5a2 + 26ab + 5b2

2. 100m2 – 240mn + 144n2

3. 8p3 + 36p2q + 54pq2 + 27q3

4. 4x4 + 23x3 + 22x2 + 49x + 70 5. 17x3 + 9y3 + 21xy2 + 30x2y 6. 9x2 – 100 7. 2a3 + 2a2b – 2ab – 2b2

8. 81 – t 4

9. a4 – a2b2 + a2 – b2

10. a3 – 3a2b + 3ab2 – b3

11. 5y3 + 20y2 + 35y + 140 12. x4 – m2

13. 27t 3 + 27t 2 + 9t + 1 14. g 2b + b2g + 5g + 5b 15. a2f + a2n + a2t 16. x2 + 11x – 180

Independent Practice, p. 160 1. 100 – 64b2

2. 8m3 – 72m2 + 216m – 216 3. g 3 + 12g 2 + 48g + 64 4. 3x6 + 4x5 + 5x4 + 12x3 + 8x2 + 10x + 12 5. 8x3 + 24x2y + 24xy2 + 8y3

6. 16a4 – 81 7. 3y3 – y2z + 3yz2 – z3 + 30y – 10z 8. 6v3w + 15w3v + 6v2 + 15w2

9. a3 – 6a2 + 12a – 8 10. x8 – y12

11. (2k – 9)3

12. x2 + 14x – 51 13. y2 – 20y + 99 14. p6 – 196 15. a4 – 256 16. x2 + 12x – 288

Progress Assessment, p. 162 1. 125x3 – 75x2y + 15xy2 – y3

2. 625 – a4

3. 7x3 + 30x2 + 29x + 6 4. 64a3 – 240a2 + 300a – 125 5. y2 – 9y – 52 6. x2 + 13x + 30

Unit 2 • operations with algebraic expressionsLesson 8: Factoring

naMe:


Assessment

Lesson Pre-AssessmentFactor the following polynomials.

1. 7p3 – 14p + 28p2

2. x2 – ax + bx – ab

3. a2 + 2ab + 3a + 6b

4. The area of a square room is x2 + 26x + 169. What is the length of each side?

5. The area of a rectangular room is x2 + 8x – 65. What might be its length?




Instruction

Lesson 8: FactoringEssential Questions 1. How is factoring related to multiplication?

2. What kind of real-life situations can require factoring?

WORDS TO KNOW



factoring to rewrite an expression as an equivalent expression that is a product

greatest common factor (GCF) in algebra, the greatest monomial that is a factor of all the terms in a polynomial or algebraic expression

monomial an expression consisting of only one term, such as 4x or 6bc


prime factor a factor that is divisible only by itself and 1



naMe:


Warm-Up Option 1Find the greatest common factor of each set of numbers.

1. 18, 24, and 48

2. 63, 99, and 81

3. 15, 105, and 135

4. 9x5 and 18x2

5. 8a3bc5, 2a6b8c2, and a6b6c


naMe:



Warm-Up Option 2Solve the following word problems by factoring.

1. The owner of a bookstore is going to close the store permanently, and wants to sell all of the remaining books in bundles. There are a total of 70 fiction books, 84 non-fiction books, and 42 art books. Each type of book must be divided evenly among the bundles. What is the greatest number of bundles that can be made with all 3 types of books?

2. Peter is filling little bags with marbles. Every bag must contain the same number of transparent marbles, the same number of silver marbles, and the same number of colored marbles. He has a total of 52 transparent marbles, 20 silver marbles, and 84 colored marbles. What is the greatest number of bags he can fill?



Instruction

Warm-Up Option 1: Debrief

1. Find the GCF of 18, 24, and 48.

The greatest common factor (GCF) of a set of two or more numbers is the largest number that can divide both numbers and result in quotients that are whole numbers.

In this case, 18, 24, and 48 can all be divided by 2, 3, and 6. The GCF of 18, 24, and 48 is therefore 6.

• In order to find the GCF of a set of numbers, express each of those numbers as a product of its prime factors:

18 = 2 • 3 • 3

24 = 2 • 2 • 2 • 3

48 = 2 • 2 • 2 • 2 • 3

• Identify the common factors as they appear in each list.

• The common factors are 2 • 3 = 6.

The GCF is 6.


• Rewrite each number as a product of its prime factors.

63 = 3 • 3 • 7

99 = 3 • 3 • 11

81 = 3 • 3 • 3 • 3


The GCF is 9.




Instruction


One of the numbers in the set, 15, is a factor of the others: both 105 and 135 can be divided by 15, and the quotient is a whole number. Therefore, the GCF is 15.

• Students can also rewrite each number as a product of its prime factors:

15 = 3 • 5

105 = 3 • 5 • 7

135 = 3 • 3 • 3 • 5


The GCF is 15.

4. Find the GCF of 9x5 and 18x2.

• First find the GCF of the coefficients.

9 = 3 • 3 • 1

18 = 3 • 3 • 2

• The GCF of 9 and 18 is 3 • 3 = 9.

• Find the GCF of the variable part of each term (x5 and x2).

• To find the GCF of the variable part, pick x with its highest exponent that divides evenly into both variables: x2

• The GCF of x5 and x2 is x2.

• Multiply the two factors, 9 and x2, to obtain the GCF, 9x2.

• You could also rewrite the expressions as products of their prime factors.

9x5 = 3 • 3 • x • x • x • x • x

18x2 = 2 • 3 • 3 • x • x

The GCF is 9x2.



Instruction

5. Find the GCF of 8a3bc5, 2a6b8c2, and a6b6c.

• First find the GCF of the coefficients.

• The coefficient of 8a3bc5 is 8, the coefficient of 2a6b8c2 is 2, and the coefficient of a6b6c is 1.

8 = 2 • 2 • 2

2 = 2

1 = 1

• The GCF of 8, 2, and 1 is 1.

• Then find the GCF of the variable part of each term.

a3bc5, a6b8c2, and a6b6c

• The GCF of the terms is a3bc.

• Multiplying these factors, we obtain the GCF:

1 • a3bc = a3bc

The GCF of 8a3bc5, 2a6b8c2, and a6b6c is a3bc.

Warm-Up Option 2: Debrief 1. The owner of a bookstore is going to close the store permanently, and wants to sell all of the

remaining books in bundles. There are a total of 70 fiction books, 84 non-fiction books, and 42 art books. Each type of book must be divided evenly among the bundles. What is the greatest number of bundles that can be made with all 3 types of books?

• To solve, find the greatest common factor (GCF) of each set of numbers. That is, find the largest number that can divide each number and result in quotients that are whole numbers.

70 = 2 • 5 • 7

84 = 2 • 2 • 3 • 7

42 = 2 • 3 • 7

• The GCF is 2 • 7 = 14.

The greatest number of bundles that can be made is 14.




Instruction

• To make the result and the situation more clear to students, ask them how many books the bundles will have of each type.

70 ÷ 14 = 5. Each bundle will have 5 fiction books.

84 ÷ 14 = 6. Each bundle will have 6 non-fiction books.

42 ÷ 14 = 3. Each bundle will have 3 art books.

2. Peter is filling little bags with marbles. Every bag must contain the same number of transparent marbles, the same number of silver marbles, and the same number of colored marbles. He has a total of 52 transparent marbles, 20 silver marbles, and 84 colored marbles. Using all the marbles, what is the greatest number of bags he can fill?

• To solve, find the GCF:

52 = 2 • 2 • 13

20 = 2 • 2 • 5

84 = 2 • 2 • 3 • 7

• The GCF is 2 • 2 = 4.

The greatest number of bags he can fill is 4.


naMe:


Focus ProblemAt a community garden meeting, the members discuss the possible measurements of each plot within the common rectangular garden. The members agree on the area of the plots each of them will have. With this information, the width and length of each plot will be determined.

1. If the agreed-upon area of each plot is 3x3 + 6x2 – 15x, what could be the length and width of each plot?

2. If the agreed-upon area of each plot is 2ax + 3bx + 14a + 21b, what could be the length and width of each plot?

3. If the agreed-upon area of each plot is x2 – 100, what could be the length and width of each plot?

4. If the agreed-upon area of each plot is x2 + 18x + 81, what could be the length and width of each plot?




Instruction


Factoring polynomials is similar to factoring numbers. In both situations, the goal is to find numbers that divide out evenly.

In the Warm-Up Options, students found the greatest common factor (GCF) of a set of numbers; that is, they were asked to find the largest number that divided evenly into each number.

• For example, to find the GCF of 30 and 105, list the prime factors of each.

30 = 5 • 3 • 2

105 = 5 • 3 • 7

The GCF is 5 • 3 = 15.

• When factoring a polynomial, the greatest common factor is the greatest monomial that is a factor of all the terms in a polynomial or algebraic expression.

• Use the following examples to illustrate the concept.

Example 1

Factor the expression 6a + 12b.

Solution

• First determine the GCF of the coefficients.

• The coefficients of 6a + 12b are 6 and 12.

• Just as before, find the prime factors of each number.

6 = 3 • 2

12 = 3 • 2 • 2

• The GCF is 3 • 2 = 6.

• Next, determine the GCF of the variables.

• The variables of 6a + 12b are a and b.

• In this case, the variables a and b do not have a common factor.

• Therefore, the GCF of 6a and 12b is 6.



Instruction

• In order to factor the expression, divide each term by 6.

6a ÷ 6 = a

12b ÷ 6 = 2b

• Rewrite the expression as a multiplication problem: one factor is the GCF and the other factor is the result of dividing each term in the expression by 6.

6a + 12b = (6)(a + 2b)

• Students may be tempted to distribute 6(a + 2b).

• Use this opportunity to point out that factoring is much like reversing the distributive property.

• If students were to distribute, the answer would be the initial expression.

• This can be used as a check, but not as a way to present the final answer.

Example 2

Factor the expression 5y5 – y2.

Solution


• The coefficients of 5y5 – y2 are 5 and –1 and have a common factor of 1.


• The variables of 5y5 – y2 are y5 and y2.

• The GCF of y5 and y2 is y2.

• Therefore, the GCF of the terms above is y2.




Instruction

• In order to factor the expression, divide each term by y2. (Refer to Lesson 6 of this unit for more information on dividing monomials.)

5y5 ÷ y2 = 5y5 – 2 = 5y3

–y2 ÷ y2 = –y2 – 2 = –y0 = –1

• Rewrite the expression as a multiplication problem: one factor is the GCF and the other factor is the result of dividing each term in the expression by y2.

5y5 – y2 = (y2)(5y3 – 1)

• A common mistake is to neglect the negative sign.

• Emphasize the importance of writing the sign correctly when dividing the second term by the GCF.


At a community garden meeting, the members discuss the possible measurements of each plot within the common rectangular garden. The members agree on the area of the plots each of them will have. With this information, the width and length of each plot will be determined.

Question 1

If the agreed-upon area of each plot is 3x3 + 6x2 – 15x, what could be the length and width of each plot?

Instruction


• The coefficients of 3x3 + 6x2 – 15x are 3, 6, and 15.

• Just as before, find the prime factors of each number.

3 = 3

6 = 3 • 2

15 = 3 • 5

• The GCF is 3.



Instruction


• The variables of 3x3 + 6x2 – 15x are x3, x2, and x.

• The GCF of x3, x2, and x is x.

• Therefore, the GCF of 3x3 + 6x2 – 15x is 3x.

• In order to factor the expression, divide each term by 3x.

3x3 ÷ 3x = x2

6x2 ÷ 3x = 2x

–15x ÷ 3x = –5

• Rewrite the expression as a multiplication problem: one factor is the GCF and the other factor is the result of dividing each term in the expression by 3x.

3x3 + 6x2 – 15x = (3x)(x2 + 2x – 5)

The length and width of plots with area 3x3 + 6x2 – 15x could be (3x) and (x2 + 2x – 5).

Question 2

If the agreed-upon area of each plot is 2ax + 3bx + 14a + 21b, what could be the length and width of each plot?

Instruction

• To find the length and width of the plot when you know its area, find numbers whose product is equal to the area, or factor the area, 2ax + 3bx + 14a + 21b.

• When approaching longer polynomials like this one, it is helpful to group terms that have common factors.

• For example, the first two terms have the common factor x, and the last two terms have the common factor 7.

2ax + 3bx + 14a + 21b

= (2ax + 3bx) + (14a + 21b)

= x(2a + 3b) + 7(2a + 3b)




Instruction

• Help students realize that the resulting expression is a binomial, and that its terms have the common factor (2a + 3b). Factor again.

x(2a + 3b) + 7(2a + 3b)

= (x + 7)(2a + 3b)

The length and the width of a plot measuring 2ax + 3bx + 14a + 21b could be (x + 7) and (2a + 3b).

• Point out to students that there is often more than one correct way to factor polynomials. In the previous example, they could have grouped the terms in a different way and still arrived at the same answer:

2ax + 3bx + 14a + 21b

= (2ax + 14a) + (3bx + 21b)

= 2a(x + 7) + 3b(x + 7)

= (x + 7)(2a + 3b)

Question 3

If the agreed-upon area of each plot is x2 – 100, what could be the length and width of each plot?

Instruction

Students may be tempted to state that this expression cannot be factored, as the terms do not have a GCF; however, this is not true.

• Notice that x2 is the square of x and 100 is the square of 10.

• In this case, the formula a2 – b2 = (a – b)(a + b) can be applied.

x2 – 100 = x2 – 102 = (x + 10)(x – 10)

The length and the width of a plot measuring x2 – 100 could be (x + 10) and (x – 10).




If the agreed-upon area of each plot is x2 + 18x + 81, what could be the length and width of each plot?

Instruction

Again, students may be tempted to state that this expression cannot be factored; however, this is not true.

• x2 + 18x + 81 is a perfect square trinomial, meaning it is the square of a binomial.

• To determine if an expression is a perfect square trinomial, examine each of the three terms.


• This is very similar to the previous question, except there is a middle term of 18x.

• The product of twice x and 9 results in the middle term of 18x.

• x2 + 18x + 81 can be written as (x + 9)2.

The length and the width of a plot measuring x2 + 18x + 81 could be (x + 9) and (x + 9).


Factor the following polynomial: (ab + 42 + 7a + 6b)

Solution

• In this example, group terms that have common factors.

(ab + 42 + 7a + 6b) = (ab + 7a) + (42 + 6b)

• Factor each group.

(ab + 7a) = a(b + 7)

(42 + 6b) = 6(7 + b)

(ab + 7a) + (42 + 6b) = a(b + 7) + 6(7 + b)

• Note that the binomials (b + 7) and (7 + b) are the same.

a(b + 7) + 6(7 + b) = (a + 6)(b + 7)





Factor the following polynomial: a2 + 4a

Solution

• Notice that a is a factor of both a2 and 4a.

a2 + 4a = a(a + 4)

Example 3

Factor the following polynomial: 9x2 + 29x + 20

Solution

• In this example, 29x can be rewritten as 9x + 20x.

• By rewriting the term 29x as two separate terms, we are able to rewrite the original expression in groups of common terms.

9x2 + 29x + 20

= 9x2 + 9x + 20x + 20

= (9x2 + 9x) + (20x + 20)

• Factor the common coefficients and variables in each binomial.

9x(x + 1) + 20(x + 1)

• Again, factor the common factor of the two terms, (x + 1).

9x(x + 1) + 20(x + 1) = (9x + 20)(x + 1)


naMe:


Guided PracticeFactor the following polynomials.

1. 2x2 – 128 6. a2 + 2a – 15

2. 125a3 + 225a2b + 135ab2 + 27b3 7. a2 – 14a – 15

3. x4k4 – 81 8. z2 + 18z + 56

4. 100 + 20e2 + e4 9. b2 + 24b + 108

5. 81 – k2 10. x2 – 5x – 24

continued


naMe:



Solve the following word problems by factoring.

11. A high school math department has a total of a2 + 13a + 42 students registered for algebra next semester, and is trying to find the optimal number of classes and students per class. What are the possible options the institute has for the number of classes and students per class?

12. Becca is trying to decide the dimensions of a pool she is designing. She knows the pool must have an area of 4m2 + 44mn + 121n2. Find a possible length and width of the pool.

13. Chang is typing an essay. He knows he can type a certain number of words per minute constantly, and in total today he typed y2 + 4y – 12 words. What might have been the number of minutes he spent typing?

14. Tia is training for a marathon. Today she ran a certain number of minutes at a certain speed (yards per minute), and in total she ran a4 – b2 yards. What might have been her speed?

15. Betsy is a drummer. Every day she plays a number of drum exercises for a certain number of minutes each. If she plays her drum exercises for a total of 5y2 – 20y minutes, what might be the number of drum exercises and minutes she plays?

16. Bali is helping build a chicken coop at a farm. He decides on the width and length, which will give the coop an area of b2 + 6b + 9. What might be the length and width of the sides?


naMe:


Independent Practice Factor the following polynomials.

1. k2 – 24k + 144 6. a2 + a – 72

2. 8x3 – 12x2 + 6x – 1 7. s 4 + 36s 2 + 99

3. 169c 8 – 121 8. x2y2 – 11xy + 28

4. 10a2 + 140ab + 490b2 9. h2 – h – 132

5. x2 + 28x + 196 10. b2 + 7b2 – 78

continued


naMe:



Solve the following word problems by factoring.

11. A theater school has a total of t 2 + t – 56 students registered for an acting class. The school is trying to decide how many separate classes to open and how many students should be in each one. What are the possible options it has for the number of classes and students per class?

12. Dr. Madden is keeping track of the heartbeats per minute of one of her patients. Today she kept track of them for a number of minutes and counted a total of n2 + 15n + 26 heartbeats. What might have been her patient’s rate of heartbeats per minute?

13. Jorge is buying red fabric to make curtains. At the store, they sell the fabric at a certain price per yard, and he buys a certain number of yards. The total amount Jorge pays is 36z2 – y2. How many yards might Jorge have bought, and what might have been the price per yard?

14. Shonda and Lucas are building a tree house in the shape of a cube and they want the volume to be w3 + 18w2 + 108w + 216 cubic feet. What should be the height of the tree house?

15. The volume of a box is 15x2 + 90x + 120. What could be its length?

16. The product of two numbers is 25x2 – 144. What are the numbers?


naMe:


Assessment

Progress AssessmentFactor the following polynomials.

1. 4x2b – 49a2b

2. 8x2 + 24x – 144

3. Sarita is writing one postcard a day. She is sending them to a certain number of her friends, and so far she has sent the same number of postcards to each. If she has sent a total of 576 – y2 postcards, what might be the number of friends she is sending them to?

4. Robert measures the length and width of a basketball court and then multiplies those dimensions to find the area. If the area of the basketball court can be represented by the polynomial ay2 + ay + 10y2 + 10y, what might have been the length and width?

5. A mechanic is keeping track of the rotations per minute of a car’s wheel to see if the wheel needs to be replaced. He counts a certain number of rotations during a certain number of minutes, with a total of k2 + 11k – 42 rotations. What might have been the wheel’s number of rotations per minute?

6. The area of a floor mat is a2 + 23a + 132. What are its length and width?




Instruction

Resource List• MathsIsFun Advanced. “Factoring in Algebra.”

www.mathsisfun.com/algebra/factoring.html

This Web site provides detailed factoring examples, with links to interactive problems students can try on their own. The interactive problems offer explanations for both correct and incorrect answers.

• Quia. “Rags to Riches.”


In this game reminiscent of “Who Wants to Be a Millionaire,” players earn “money” by choosing the correct answer from four choices. Three hints are offered for each question. Incorrect answers result in the loss of the game.

• Seward, Kim, for West Texas A&M University Virtual Math Lab. “Intermediate Algebra Tutorial 28: Factoring Trinomials.”

www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut28_facttri.htm

This Web site provides a step-by-step process for factoring trinomials, with examples.



Lesson 8 Answer KeyLesson Pre-Assessment, p. 165 1. 7p(p2 – 2 + 4p) 4. (x + 13) 2. (x – a)(x + b) 5. (x + 13) or (x – 5) 3. (a + 2b)(a + 3)

Guided Practice, p. 181 1. (2)(x + 8)(x – 8) 2. (5a + 3b)3

3. (x2k2 + 9)(xk + 3)(xk – 3) 4. (e 2 + 10)2

5. (9 – k)(9 + k) 6. (a + 5)(a – 3) 7. (a – 15)(a + 1) 8. (z + 14)(z + 4) 9. (b + 18)(b + 6) 10. (x – 8)(x + 3) 11. (a + 6) and (a + 7) 12. (2m + 11n) and (2m + 11n) 13. (y + 6) or (y – 2) 14. (a2 – b) or (a2 + b) 15. (5y) or (y – 4) 16. (b + 3) and (b + 3)

Independent Practice, p. 183 1. (k – 12)2 9. (h + 11)(h – 12) 2. (2x – 1)3 10. (b + 13)(b – 6) 3. (13c4 – 11)(13c4 + 11) 11. (t + 8) and (t – 7) 4. (10)(a + 7b)2 12. (n + 13) or (n + 2) 5. (x + 14)2 13. (6z – y) or (6z + y) 6. (a + 9)(a – 8) 14. (w + 6) 7. (s2 + 3)(s2 + 33) 15. (x + 4) or (x + 2) or 15 8. (xy – 7)(xy – 4) 16. (5x + 12) and (5x – 12)

Progress Assessment, p. 185 1. (b)(2x – 7a)(2x + 7a) 4. (a + 10) or (y2 + y) 2. (8)(x + 6)(x – 3) 5. (k + 14)(k – 3) 3. (24 – y) or (24 + y) 6. (a + 12)(a + 11)

Unit 2 • operations with algebraic expressionsLesson 9: Simplifying Algebraic Fractions

naMe:



Assessment

Lesson Pre-AssessmentSimplify the following expressions, noting the domain.

1. 8 2 4

2

3 2x x xx

+ −

2. a a

a

2

2

4 44

− +−

3. x x

x

2 20 10010 100+ +

+

4. x xx x

2

2

23 13226 165

+ ++ +

5. The area of a rectangle is x2 + 20x + 75 centimeters. If the length of the rectangle is (x + 5), what is its width?



Instruction

Lesson 9: Simplifying Algebraic FractionsEssential Questions 1. How is simplifying algebraic fractions similar to simplifying numerical fractions?

2. How can you determine the domain restrictions of a rational expression?

WORDS TO KNOW

cancelling simplifying a rational expression by eliminating factors that the numerator and denominator have in common



domain the values that x can take within a certain rational expression

domain restrictions the value(s) of x that are excluded from the domain because they make the rational expression undefined





rational expression a fraction with algebraic expressions in the numerator and denominator

simplest form a rational expression whose common factors between the numerator and denominator have been canceled and cannot be simplified any further

undefined a rational expression whose denominator is equal to 0



naMe:



Warm-Up Option 1Simplify the following fractions.

1. 2540

2. 5 2

4

xx

3. 69

4 2 7

2 6 3

a b ca b c

Factor the following expressions.

4. 4a2 + 8ab + 4b2

5. x2 – 7x – 30

6. y2 – 9


naMe:


Warm-Up Option 2Use factoring to solve the following problems.

1. A rectangular rug has an area of x2 – 4x – 77 inches. What could be the length of the rug?

2. A rectangular porch has an area of (x2 – 169). What could be the length and width of the porch?




Instruction

Warm-Up Option 1: DebriefSimplify the following fractions.

1. 2540

• To simplify the fraction, find the common factors of the numerator and denominator.

2540

5 55 8

=••

• Cancel the factors that the numerator and denominator have in common.

2540

5 55 8

58

=••

=

2540

58

=

2. 5 2

4

xx

• To simplify, find the common factors of the numerator and denominator.

5 52

4

2

2 2

xx

xx x

=••

• Cancel the common factor of x2.

5 5 52

4

2

2 2 2

xx

x

x x x=

••

=

5 52

4 2

xx x

=



Instruction

3. 69

4 2 7

2 6 3

a b ca b c

• To simplify, find the common factors of the numerator and denominator.

69

2 33 3

4 2 7

2 6 3

2 2 2 3 4

2 2 4

a b ca b c

a a b c ca b b c

=• • • • • •• • • • • 33

• Cancel the common factors.

69

2 3

3 3

4 2 7

2 6 3

2 2 2 3 4

2 2 4

a b ca b c

a a b c c

a b b c=

• • • • • •• • • • • 33

2 4

4

23

=a cb

69

23

4 2 7

2 6 3

2 4

4

a b ca b c

a cb

=

4. 4a2 + 8ab + 4b2

• Each term of the expression 4a2 + 8ab + 4b2 can be factored by 4.

4a2 + 8ab + 4b2 = 4(a2 + 2ab + b2)

• (a2 + 2ab + b2) is the square of a binomial.

• Notice that a2 is the square of a, and b2 is the square of b.

• The product of twice a and b is 2ab, the middle term of the polynomial.

• (a2 + 2ab + b2) can be written as (a + b)2.

4a2 + 8ab + 4b2 = 4(a + b)2

5. x2 – 7x – 30

• To factor x2 – 7x – 30, find two numbers whose product is –30 and whose sum is –7.

• The numbers –10 and +3, when multiplied, result in –30; when added, they result in –7.

The factored form of x2 – 7x – 30 is (x – 10)(x + 3).




Instruction

6. y2 – 9

• Notice that y2 is the square of y and 9 is the square of 3.


y2 – 9 = y2 – 32 = (y + 3)(y – 3)

The factored form of y2 – 9 is (y + 3)(y – 3).

Warm-Up Option 2: Debrief 1. A rectangular rug has an area of x2 – 4x – 77 inches. What could be the length of the rug?

• Refer to the area formula for a rectangle:

Area = length • width = l • w

x2 – 4x – 77 = l • w

• To solve, find the factors of the area x2 – 4x – 77.

• To factor x2 – 4x – 77, find two numbers whose product is –77 and whose sum is –4.

• The numbers –11 and 7, when multiplied, result in –77; when added, they result in –4.

• The factored form of x2 – 4x – 77 is (x – 11)(x + 7).

The length of the rug could be either of the factors (x + 7) or (x – 11).

2. A rectangular porch has an area of (x2 – 169). What could be the length and width of the porch?

• Refer to the area formula for a rectangle:

Area = length • width = l • w

(x2 – 169) = l • w

• To solve, find the factors of the area (x2 – 169).



x2 – 169 = x2 – 132 = (x + 13)(x – 13)

The length and width of the porch could be (x – 13) and (x + 13).


naMe:


Focus ProblemA circus school is opening and x2 + x – 56 students have registered for its first semester. The headmaster is still trying to figure out how to divide up the students into different classes.

1. Find the expression for the number of students per class in simplest form if the headmaster decides to divide the students into x2 + 9x + 8 classes.

2. Find the expression for the number of students per class in simplest form if the headmaster decides to divide the students into x2 – 6x – 7 classes.

3. At the last minute, another 26 students register. Now the headmaster decides to divide the total number of students into x2 – 10x + 25 classes. Find the number of students there will be per class.




Instruction


Fractions in algebra are often called rational expressions, or fractions with algebraic expressions in the numerator and denominator.

• To write a rational expression in its simplest form, divide both the numerator and denominator by the same term. When all factors have been cancelled from the numerator and denominator, the simplest form remains.

• Rational expressions often have domain restrictions, or the value(s) of x that are excluded from the domain because they make the rational expression undefined. An undefined rational expression has a denominator of 0.

• Students often struggle with the concept of an undefined denominator. Remind students that it is impossible to divide a group of 16 students into groups of 0. This is similar to the division of 16 by 0; it is not possible. Many calculators return a divide by 0 error to the user when this calculation is attempted.

• Use the following example to illustrate simplifying algebraic fractions and finding the domain of the expression.

Example

Simplify the expression x x

x

2

2

8 1616

+ +−

. Then find the domain of x.

Solution

• To simplify a rational expression, begin by factoring both the numerator and denominator. (Refer to Lesson 8 of this unit for more information on factoring polynomials.)

x xx

xx x

2

2

28 1616

44 4

+ +−

=+

+ −( )

( )( )

• Cancel common factors. In this case, the only factor common to both the numerator and denominator is (x – 4).

( )( )( )

( )( )( )( )

( ) (xx x

x xx x

x x++ −

=+ ++ −

=+ +4

4 44 44 4

4 42 ))

( ) ( )

( )( )x x

xx+ −

=+−4 4

44



Instruction

• The domain, or value for x, is restricted when any of the factors is equal to 0, causing the product to be 0.

• Students often neglect to look at all the factors of the denominator, focusing primarily on the simplified form. This error will not provide all restrictions.

• Be sure students understand that “all the factors of the denominator” refers to the factored fraction before common factors are cancelled.

• (x – 4) is equal to 0 when x is equal to 4.

• (x + 4) is equal to 0 when x is equal to –4.

• The domain is restricted to x ≠ –4 and x ≠ 4.

x xx

xx

2

2

8 1616

44

+ +−

=+−

( )( )

, but x cannot equal 4 or –4.


A circus school is opening and x2 + x – 56 students have registered for its first semester. The headmaster is still trying to figure out how to divide up the students into different classes.

Question 1

Find the expression for the number of students per class in simplest form if the headmaster decides to divide the students into x2 + 9x + 8 classes.

Instruction

• To find the number of students per class, divide the total number of students by the number of classes.

• Set up a fraction and reduce it to its simplest form.

• Total number of students: x2 + x – 56

• Total number of classes: x2 + 9x + 8

x xx x

2

2

569 8

+ −+ +




Instruction

• Begin by factoring both the numerator and denominator.

x xx x

x xx x

2

2

569 8

8 78 1

+ −+ +

=+ −+ +

( )( )( )( )

• Explain to students that a rational expression is written in simplest form when all common factors of the numerator and denominator have been cancelled, so that it cannot be simplified any further.

• Simplify.

( ) ( )

( ) ( )

( )( )

x x

x xxx

+ −+ +

=−+

8 7

8 1

71

• To determine the domain restrictions, find the values of x in the denominator that result in a factor of 0.

• Again, be sure to consider all factors of the denominator.



• Therefore, the domain is restricted to x ≠ –8 and x ≠ –1.

The number of students per class is ( )( )

xx−+71

, but x cannot equal –8 or –1.

Question 2

Find the expression for the number of students per class in simplest form if the headmaster decides to divide the students into x2 – 6x – 7 classes.

Instruction




• Total number of classes: x2 – 6x – 7

x xx x

2

2

566 7

+ −− −



Instruction


x xx x

x xx x

2

2

566 7

8 77 1

+ −− −

=+ −− +

( )( )( )( )

• Simplify.

( ) ( )

( ) ( )

( )( )

x x

x xxx

+ −− +

=++

8 7

7 1

81


• Again, be sure to consider all factors of the denominator.



• Therefore, the domain is restricted to x ≠ 7 and x ≠ –1.


xx

++81

, but x cannot equal 7 or –1.

Question 3

At the last minute, another 26 students register. Now the headmaster decides to divide the total number of students into x2 – 10x + 25 classes. Find the number of students there will be per class.

Instruction

• The original number of registered students was x2 + x – 56. Now an additional 26 have registered, so we have to add them to find the new number of registered students.

x2 + x – 56 + 26 = x2 + x – 30





Instruction



• Total number of classes: x2 – 10x + 25

x xx x

2

2

3010 25+ −− +


x xx x

x xx

2

2 2

3010 25

6 55

+ −− +

=+ −

−( )( )

( )

• Simplify.

( )( )( )

( ) ( )

( ) ( )

( )(

x xx

x x

x xxx

+ −−

=+ −− −

=+−

6 55

6 5

5 5

652 ))



• Therefore, the domain is restricted to x ≠ 5.


xx+−65

, but x cannot equal 5.


Find the domain of the following rational expression: xx

3

2

13 3

+−

Solution

• To find the domain of the expression, we must find the values that x cannot be.

• x cannot be any value that makes the denominator equal to 0.

• Begin by factoring the denominator.

• 3 can be factored from both terms.

3x2 – 3 = 3(x2 – 1)



Instruction

• Notice that x2 is the square of x, and 1 is the square of 1.


x2 – 1 = x2 – 12 = (x + 1)(x – 1)

• The denominator can be rewritten as 3(x – 1)(x + 1).

• The denominator is restricted when any of the factors is equal to 0, causing the product to be 0.



• The domain restrictions for xx

3

2

13 3

+−

are x ≠ –1, x ≠ 1.

Therefore, the domain is all real numbers except x = –1 or x = 1.

Example 2

Simplify 3 4810 24

2

2

xx x

−− +

and note the domain restrictions.

Solution

• To simplify a rational expression, begin by factoring both the numerator and denominator.

3 4810 24

3 16

4 63 4 42

2

2xx x

xx x

x x−− +

=−

− −=

+ −( )

( )( )( )( )(xx x− −4 6)( )

• Cancel common factors. In this case, the only factor common to both the numerator and denominator is (x – 4).

3 4 4

4 6

3 46

( ) ( )

( ) ( )

( )( )

x x

x xxx

+ −− −

=+−

3 4810 24

3 46

2

2

xx x

xx

−− +

=+−

( )( )




Instruction

• To determine the domain restrictions, refer to all the factors of the original denominator, (x – 4) and (x – 6).

• When x = 4 or x = 6, the denominator would be 0.

• Therefore, the domain restriction is x ≠ 4, x ≠ 6.

3 4810 24

3 46

2

2

xx x

xx

−− +

=+−

( )( )

, x ≠ 4, x ≠ 6


naMe:


Guided PracticeFind the domain of the following rational expressions.

1. 12

24 1442x x− +

2. ab x

x ax bx ab−

+ + +2

3. x x x

x

3 2

2

1196

+ + +−

4. z

z z2 5 104− −

5. ( )( )( )x x x

x− − −1 2 3

Reduce the following rational expressions to their simplest form.

6. 2 128

16 64

2

2

xx x

−− +

7. x x x

x x

3 2

2

18 108 2162 24 72

+ + ++ +

8. x xy x y

x

2

2

3 3

9

+ + +−

9. ab a b

b b− +

− −6 4 24

5 62

–

10. x x

x x

2

2

15 4411

+ ++

continued


naMe:



Solve the following problems by writing the rational expressions in their simplest form.

11. Claudia is typing an article for her school newspaper. She has typed x2 – 3x – 10 pages in x2 – 2x – 8 minutes. How many pages per minute has she typed?

12. Zack is organizing a raffle. He has a total of x2 – 8x + 7 raffle tickets to distribute evenly among x2 – 17x + 70 of his friends, who are going to sell them. How many tickets does each friend have to sell?

13. The area of a tennis court is x2 + 22x + 121, and its length is x2 – 121. What is its width?

14. In a cup stacking contest, Max stacked x3 + 12x2 + 48x + 64 cups in a period of 2x2 + 16x + 32 seconds. How many cups per second did Max stack?

15. At the Olympics, the gold medal winner of the javelin throw competition threw the javelin a distance of x2 – 4x – 77 meters. The javelin flew in the air for x2 + x – 42 seconds before falling on the ground. What was the speed of the javelin?

16. Preparing for a race, Marta ran a distance of x2 – 169 meters in x2 + 19x + 78 seconds. What was Marta’s speed?


naMe:


Independent Practice Find the domain of the following rational expressions.

1. 3

3 30 75

2

2

xx x− +

2. b c

b b b

3 3

3 22 18 54 54−

− + −

3. 22

1322x x+ −

4. z

z−−50

5 1252

5. x x x x

x

4 3 2

4

3 6 9 12+ + + +

Reduce the following rational expressions to their simplest form.

6. a a a

a a

3 2

2

3 6 1817 42

− +− +

–

7. 4 40 36

2 162

2

2

x xx− +

−

8. 3 9 9 39 18 9

3 2

2

x x xx x− + −− +

9. a ab a b

a a

2

2

11 113 24 99− + −

+ −

10. b b

b b

2

3 2

11 284

− +−

continued


naMe:



Solve the following problems by writing the rational expressions in their simplest form.

11. A bedroom has an area of x2 + 14x + 49 square feet and a length of 3x + 21 feet. What is the width of the bedroom?

12. A bakery uses a total of x2 – 18x + 80 pounds of bread dough to make x2 – 14x + 48 loaves of bread. How many pounds of dough are used to make each loaf?

13. Carly is training for a swimming race. She swims x2 + 4x – 45 laps in x2 + 18x + 81 minutes. How many laps per minute does she swim?

14. Sam is an urban planner for the city. She determines that in one day a total of x2 + 21x – 162 passengers use one of the x – 6 city buses. How many passengers use each bus?

15. Josh has x2 + 9x + 14 cookies and wants to distribute them evenly among his x2 – 4x – 91 friends. How many cookies will each friend get?

16. A dog runs x2 + 25x – 26 meters in x2 + 9x – 10 seconds. How fast does the dog run in meters per second?


naMe:


Assessment

Progress AssessmentSimplify the following expressions, noting the domain.

1. x x

x x

2

2

8 10530 225

− −− +

2. x x

x

2

2

34 225625

+ +−

Solve the following problems by simplifying rational expressions.

3. At a restaurant, x2 + 14x – 51 customers are sitting at x2 + 10x – 39 tables. How many customers are there on average per table?

4. At the same restaurant, Peter used x3 + 6x2 + 12x + 8 pounds of strawberries to make x2 – 4 pies. How many pounds did he use for each pie on average?

5. Lena is training a horse. The horse can run a distance of x2 + 12x + 36 meters in x2 – 6x – 72 seconds. What is the speed of the horse in meters per second?

6. The area of a baking sheet is x2 + 16x + 28 square centimeters. If the length of the baking sheet is x + 14 centimeters, what is its width?




Instruction

Resource List• Dawkins, Paul, for Paul’s Online Math Notes. “Algebra: Rational Expressions.”

http://tutorial.math.lamar.edu/Classes/Alg/RationalExpressions.aspx

This E-book provides an exhaustive tutorial on simplifying rational expressions, with limited information on domain restrictions.

• Dendane, A. “Simplify Rational Expressions.”

www.analyzemath.com/Rational_expressions/Simp_rat_expre.html

This Web site provides a good introduction to how to simplify rational expressions, with examples and detailed solutions.

• Stapel, Elizabeth, for Purplemath. “Polynomial Division: Simplification and Reduction.”

www.purplemath.com/modules/polydiv.htm

This site explains in detail how to simplify rational expressions. The tutorial is three pages long, but only the first one deals with simplification; the second page moves on to long division of polynomials, which is the subject matter of Lesson 10.

• Stapel, Elizabeth, for Purplemath. “Rational Expressions: Finding the Domain.”

www.purplemath.com/modules/rtnldefs.htm

This site explains in detail how to identify domain restrictions and find the domain of a rational expression.



Lesson 9 Answer KeyLesson Pre-Assessment, p. 188

1. 4x2 + x – 2, x ≠ 0 4. ( )( )

xx

++1215

, x ≠ –15, –11

2. aa−+

22

, a ≠ –2, 2 5. (x + 15)

3. x + 1010

, x ≠ –10

Guided Practice, p. 203 1. all real numbers except x = 12

2. all real numbers except x = –a, –b

3. all real numbers except x = –14, 14


5. all real numbers except x = 0

6. 2 8

8( )( )

xx+−

, x ≠ 8

7. ( )x + 6

2, x ≠ –6

8. ( )( )x yx+− 3

, x ≠ –3, 3

9. ( )( )ab

++41

, b ≠ –1, 6

10. ( )x

x+ 4

, x ≠ –11, 0

11. ( )( )

xx

−−54

, x ≠ –2, 4

12. ( )( )

xx

−−

110

, x ≠ 7, 10

13. ( )( )

xx+−1111

, x ≠ –11, 11

14. ( )x + 4

2, x ≠ –4

15. ( )( )xx−−116

, where x ≠ –7, 6

16. ( )( )xx−+136

, where x ≠ –6, –13

Independent Practice, p. 205 1. all real numbers except x = 5





6. ( )( – )aa

2 614

+, a ≠ 3, 14

7. 2 1

9( )( )

xx−+

, x ≠ –9, 9

8. ( )x − 1

3, x ≠ 1

9. ( – )( )a ba3 3−

, a ≠ –11, 3

10. ( )b

b− 72 , b ≠ 0, 4

11. ( )x + 7

3, x ≠ –7

12. ( )( )xx−−106

, x ≠ 6, 8

13. ( )( )

xx−+59

, x ≠ –9

14. (x + 27), x ≠ 6

15. ( )( )

xx+−

213

, x ≠ –7, 13

16. ( )( )xx

++2610

, x ≠ –10, 1


1. ( )( )

xx+−

715

, x ≠ 15 4. ( )( )xx+−22

2

, x ≠ –2, 2

2. ( )( )

xx+−

925

, x ≠ –25, 25 5. ( )( )

xx+−

612

, x ≠ –6, 12

3. xx

++1713

, x ≠ –13, 3 6. (x + 2), x ≠ –14

Unit 2 • operations with algebraic expressionsLesson 10: Dividing Polynomials

naMe:



Assessment

Lesson Pre-AssessmentDivide the following polynomials.

1. (x2 – 5x + 5) ÷ (x – 1)

2. (x2 + 6x + 44) ÷ (x + 5)

3. (x2 + 18x + 20) ÷ (x + 1)

4. Jena is training for a swimming competition. She can swim a distance of x2 + 3x + 12 meters in x + 4 seconds. How many meters does she swim per second?

5. At a restaurant, x2 – 10x – 50 customers are sitting at x – 3 tables. How many customers are there on average per table?



Instruction

Lesson 10: Dividing PolynomialsEssential Questions 1. How is dividing polynomials similar to dividing real numbers?

2. How can we apply division to real-world situations?

3. What do the quotient and remainder mean in a real-world situation?

WORDS TO KNOW




dividend in a division problem, the number that is the whole divided in parts

divisor in a division problem, the number that divides the dividend






quotient the result of division


remainder in a division problem, the portion of the dividend that does not divide exactly into the divisor, and that is left after dividing




naMe:



Warm-Up Option 1Find each quotient. Express your answer using remainders.

1. 47 ÷ 2

2. 362 ÷ 24

3. (145 + x) ÷ 12

4. (6x + 13) ÷ 6

5. (x2 + 17x + 14) ÷ x


naMe:


Warm-Up Option 2Solve the following division problems. Express your answers using remainders.

1. The area of a house is (258 + x) square feet. The house is 24 feet long. What is its width?

2. David is training for a track meet. He can run a distance of 3x + 14 meters in x seconds. How many meters does he run per second?




Instruction

Warm-Up Option 1: Debrief 1. 47 ÷ 2

• 47 divided by 2 does not result in a whole number.

• In order to perform this division, set up a long division problem.

)2 47

• Remind students that the number being divided, 47, is called the dividend.

• The number that we divide by, 2, is called the divisor.

• The result of a division problem is called the quotient.

• When a quotient does not go exactly into a dividend, any number left over is called the remainder.

• Long division is performed digit by digit.

• Divide the first digit of the dividend by the divisor.

4 ÷ 2 = 2

• Put this first digit of the quotient, 2, above the digit 4 that you just divided.

)2 472

• Multiply the quotient by the divisor and put the result under the digit you divided in the dividend.

• Subtract the result from the first digit in the dividend.

)2 47

4

2

0

−



Instruction

• Now bring down the next digit of the dividend.

)2 4 7

4

07

2

↓

−

• This is the new part of the dividend that you will divide now.

• Divide again by the divisor.

7 ÷ 2 = 3, remainder 1

• Again, put the result of this division as the next digit in the quotient.

)2 4 7

4

07

23

↓

−

• Multiply that last digit of the quotient by the divisor.

3 • 2 = 6

• Put that number under the part of the dividend you are now working with, 07, and subtract again.

)2 4 7

4

07

6

1

23

↓

−

−

• You have no more digits in the dividend to divide. This is the end of the long division problem.

The quotient is 23 with a remainder of 1.




Instruction

In previous units, division was continued until a decimal pattern was discovered, or the decimal terminated. In this lesson, students will be asked to identify the quotient with a remainder.

• Since we have a divisor of 2 and a remainder of 1, 23 remainder 1 can also be expressed as

2312

.

2. 362 ÷ 24

• Set up a long division problem and solve.

)24 36215

24

122

120

2

−

−

The quotient is 15 remainder 2.

• 15 remainder 2 can also be expressed as 15224

, where 2 is the remainder and 24 is the

divisor. This reduces to 15112

.

3. (145 + x) ÷ 12

• 145 + x is a polynomial. The procedure to divide it by 12 is not very different from the long division for real numbers of the previous examples.


)12 14512

12

25

24

1

+

−

−

x

++ x



Instruction

• When you get to this point, (1 + x) is not divisible by 12. Treat this as your remainder.

The quotient of 145 + x and 12 is 12 with remainder 1 + x, which can be expressed as 12112

++ x

.

4. (6x +13) ÷ 6


)6 6 132

6

13

12

xx

x

++

−+−++1

• The quotient of (6x +13) and 6 is x + 2 with remainder 1, which can be expressed as x + 216

.

5. (x2 + 17x +14) ÷ x

• This problem looks more complicated, but is essentially the same.


)x x xx

x

x

2

2

17 1417

17

+ ++

−+

+

−1714

x

14 is not divisible by x. Therefore, the remainder is 14. The quotient is x + 17.

The quotient of (x2 + 17x +14) ÷ x can be expressed as xx

+ 1714

.




Instruction

Warm-Up Option 2: Debrief 1. The area of a house is 258 + x square feet. The house is 24 feet long. What is its width?

• 258 + x is a polynomial. The procedure to divide 258 + x by 24 is not very different from the long division of real numbers.


• First determine how many times 24 goes into 258.

)24 25810

240

0

1

+

−

−

x

18

88 + x

• When you get to this point, (18 + x) is not divisible by 24. Treat this as your remainder.

The quotient of 258 + x and 24 is 10 and the remainder is 18 + x, which can be expressed as

101824

++ x

, where 18 + x is the remainder and 24 is the divisor.

2. David is training for a track meet. He can run a distance of 3x + 14 meters in x seconds. How many meters does he run per second?

• To divide the polynomial 3x + 14 by x, follow the same procedure as in the previous problem.


)x x

x

3 143

3

14

+

−+

• The remainder is 14, because it is not divisible by x.

The quotient of 3x + 14 divided by x is 3 with a remainder of 14, which can be expressed as 3 + 314x

.


naMe:


Focus ProblemSylvia has x3 + 18x2 + 18x + 24 salmon on her fishing boat. She wants to put all the salmon in a number of containers to handle them more easily.

1. If Sylvia wants to store the same number of salmon in each of x + 8 containers, how many salmon will she have to put in each container to hold all the salmon that she caught?

2. Interpret the results of question 1. How many salmon will she have to put in how many containers? Will there be any salmon left over?

3. Sylvia decides that only x + 1 salmon fit in each container. How many containers will she need to hold all the salmon?

4. If Sylvia buys containers that hold x + 17 salmon, how many containers will she need?




Instruction


Solving polynomial divisions is not that different from long division of whole numbers.

• First, set up the long division.

• Then, take one term of the dividend at a time to generate one term of the quotient at a time.

• Proceed as with regular long division, multiplying the latest term in the quotient by the dividend, and subtracting the result from the dividend.

• Bring down the next term of the dividend, and find the next term of the quotient.

• Repeat the process until you have divided the whole dividend. You will obtain a quotient and possibly a remainder, just as in long division of whole numbers.

• Use the following example to illustrate this procedure.

Example

Divide (4x2 + 18x – 24) by (x – 2).

Solution

• Set up the long division.

)x x x

x x

− + −

− +

2 4 18 24

4 8

2

2

• Divide the first term in the dividend by the first term in the divisor.

4x2 ÷ x = 4x

)x x x

x

x− + −

−

2 4 18 24

4

42

2

• 4x will be the first term in the quotient.

• Multiply 4x by the dividend.

(4x)(x – 2) = 4x2 – 8x



Instruction

• Write 4x2 – 8x under the dividend with the opposite signs (–4x2 + 8x) in order to subtract it.

)x x x

x x

x− + −

− +

2 4 18 24

4 8

42

2

• Students may forget that they are subtracting all of (4x2 + 8x) from (4x2 + 18x). When subtracting (4x2 + 8x), each term must be subtracted, not just 4x2.

)x x x

x x

x

x− + −

− ++

2 4 18 24

4 8

0 26

42

2

• The result is 26x.

• Bring down the next term in the dividend to obtain 26x – 24.

• Repeat the process. Divide the first term by the first term in the divisor to obtain the next term in the quotient: 26x ÷ x = 26

x x x

x x

x

x− + −

− +−

+2 4 18 24

4 8

24

4 262

2

26

)

• Now multiply that new term of the quotient by the divisor.

26 • (x – 2) = 26x – 52




Instruction

• Subtract this result from the last part of the dividend you were using. Reverse the signs of the result in order to subtract:

26x – 52 becomes –26x + 52

x x x

x x

x

x

− + −

− +−

− +

2 4 18 24

4 8

24

26

2

2

26

)

552

28

4 26x +

• The difference is 28.

• Since there are no more terms in the dividend to use, we have completed the division problem.

The quotient of (4x2 + 18x – 24) and (x – 2) is 4x + 26 with a remainder of 28. The answer can be

expressed as 4 2628

2x

x+

− + 4 26

282

xx

+−

+ 4 2628

2x

x+

−, where 28 is the remainder and x – 2 is the divisor.


Sylvia has x3 + 18x2 + 18x + 24 salmon on her fishing boat. She wants to put all the salmon in a number of containers to handle them more easily.

Question 1

If Sylvia wants to store the same number of salmon in each of x + 8 containers, how many salmon will she have to put in each container to hold all the salmon that she caught?

Instruction

• To solve, set up a long division problem for (x3 + 18x2 + 18x + 24) ÷ (x + 8).



Instruction

• Follow the processes previously described to yield the resulting equation:

x x x x

x x

x x

x

+ + + +

− −

+

−

8 18 18 24

8

10 18

10

3 2

3 2

2

2 −−− +

+

80

62 24

62 496

x

x

x

)

520

10 622x x+ −

• She needs to put x2 + 10x – 62 in each container, but she will have a remainder of 520.

Question 2

Interpret the results of question 1. How many salmon will Sylvia have to put in how many containers? Will there be any salmon left over?

Instruction

• The dividend, x3 + 18x2 + 18x + 24, is the total number of salmon she caught.

• The divisor, x + 8, is the number of containers she wants to use.

• The quotient, x2 + 10x – 62, is the number of salmon per container.

• The remainder, 520, is salmon left over.

Sylvia will have to put x2 + 10x – 62 salmon in each of x + 8 containers if she wants them all to hold the same number of salmon. However, there will be 520 salmon left over.

Question 3

Sylvia decides that only x + 1 salmon fit in each container. How many containers will she need to hold all the salmon?

Instruction

• To solve, set up a long division problem for (x3 + 18x2 + 18x + 24) ÷ (x + 1).




Instruction

• Follow the processes previously described to yield the resulting equation:

x x x x

x x

x x

x

+ + + +

− −

+

− −

1 18 18 24

17 18

17

3 2

3 2

2

2 117

24

x

x +

)

−− −

+ +

x

x x

1

23

17 12

She will need x2 + 17x + 1 containers, but she will have 23 salmon left over.

Question 4

If Sylvia buys containers that hold x + 17 salmon, how many containers will she need?

Instruction

• Set up the new long division and solve.

x x x x

x x

x

+ + + +

− −

+

17 18 18 24

17

18

3 2

3 2

2 xx

x x

x

− −+

2 17

24

− −x 17

)

7

x x2 1+ +

• The answer is x xx

2 1717

+ + ++

.

Sylvia now needs x2 + x + 1 containers, and she will have 7 salmon left over.



Instruction


Divide x2 + 7x + 90 by x – 3.

Solution


x x x

x x

x

x

− + +

− ++

− +

3 7 90

3

90

10 30

2

2

10

)

120

x + 10

The quotient of x2 + 7x + 90 and x – 3 is xx

+ +−

10120

3.

Example 2

Divide x2 + 7x + 90 by x + 3.

Solution


x x x

x x

x

x

+ + +

− −+

− −

3 7 90

3

90

4 12

2

2

4

)

78

x + 4

The quotient of x2 + 7x + 90 and x – 3 is xx

+ ++

478

3.


naMe:



Guided PracticeDivide the following polynomials.

1. (x2 – 20x + 105) ÷ (x – 9)

2. (x2 + 36x + 108) ÷ (x + 18)

3. (x3 – 15x2 + 100x – 120) ÷ (x – 4)

4. (11x3 – 5x2 – 45x – 76) ÷ (x + 1)

5. (x3 – 4x2 – 8x + 99) ÷ (x – 6)

6. (x4 + 6x3 + 5x2 – 30x – 30) ÷ (x + 4)

7. (2x4 – 8x3 – 2x2 + 10x + 50) ÷ (x – 5)

8. (5x4 + 10x3 – 40x2 + 130x – 150) ÷ (x + 5)

9. (x4 – 12x3 – 14x2 + 300x – 120) ÷ (x – 6)

10. (x4 + 8x3 – 5x2 – 2x + 32) ÷ (x – 2)

continued


naMe:


Solve the following word problems by dividing polynomials.

11. At Angelo’s bologna factory, a total of 2x2 – 20x + 80 pounds of meat are processed each day into x – 8 loaves of bologna. How many pounds of meat does each loaf of bologna have?

12. Amanda is competing on a game show. She correctly answers x3 + x2 – 50x + 250 questions in x + 10 minutes. How many questions per minute does she answer?

13. Leah is managing an art museum. She determines that an average of x2 + 11x – 152 patrons visit the museum every x – 6 days. How many patrons on average visit the museum each day?

14. A restaurant has x2 – 3x + 36 ounces of soup, and serves it in equal parts in x – 4 bowls. How many ounces of soup were in each bowl?

15. A city block has an area of x2 + 6x + 88 square meters. If the city block is x + 4 meters wide, how long is it?

16. A pool has a depth of x + 6 feet and a volume of x3 + 7x2 + 10x + 34 cubic feet. What is the area of the pool?


naMe:



Independent Practice Divide the following polynomials.

1. (x2 – 2x + 10) ÷ (x + 1)

2. (x2 + 7x – 108) ÷ (x – 10)

3. (x3 – 35x2 + 135x – 350) ÷ (x – 7)

4. (10x3 – 5x2 – 75x + 255) ÷ (x + 5)

5. (x3 + 5x2 + 5x + 90) ÷ (x + 6)

6. (3x4 + 9x3 + 27x2 – 81x – 243) ÷ (x + 3)

7. (4x4 – 8x3 – 20x2 + 40x + 120) ÷ (x – 4)

8. (x4 + 9x3 – 30x2 + 128x – 330) ÷ (x – 3)

9. (x4 + 11x3 – 17x2 – 41x – 111) ÷ (x + 7)

10. (4x4 – x3 + 5x2 – 9x + 25) ÷ (x – 1)

continued


naMe:


Solve the following word problems by dividing polynomials.

11. The area of a city park is x2 + 22x + 220 and its length is x + 11. What is its width?

12. Claudia is typing a novel. She has typed x2 – 3x – 10 pages in x + 3 days. How many pages per day did she type?

13. At a large day care, there are x3 + 15x2 – 50x + 350 toddlers in x + 5 classrooms. What is the average number of toddlers per classroom?

14. Terrence is organizing a voter registration drive. He has a total of x3 – x2 – 8x + 55 blank registration cards to distribute evenly among his x + 4 volunteers. How many registration cards does each volunteer get, and how many will Terrence have left over?

15. Alberto has x2 – 4x – 40 chocolate bars that he divides among x + 5 friends. How many chocolate bars does each friend receive on average?

16. A shopping Web site reports that 2x3 + x2 – 6x – 32 dollars were spent in one day by x – 2 customers. What was the average amount that each customer spent?


naMe:



Assessment

Progress AssessmentDivide the following polynomials.

1. (x3 + x2 – 8x – 64) ÷ (x + 4)

2. (x3 + 10x2 + 10x + 4) ÷ (x + 8)

3. A baseball player threw the ball across the field a distance of x2 – 8x – 24 meters. The ball flew in the air for x – 4 seconds before falling on the ground. What was the speed of the ball?

4. Andrew is a sports car test driver. On a test track, he was able to drive a new coupe x2 + 30x – 196 miles in x + 14 seconds. What was Andrew’s speed?

5. It’s lasagna day at the school cafeteria, and a total of x3 + 4x2 + x + 144 pounds of lasagna have been prepared. If each serving of lasagna weighs x + 9 pounds, how many servings of lasagna can be sold?

6. Bree has 7x3 + 15x2 + 9x + 30 truffles. She decides to make presents by filling small boxes with the same amount of truffles each. If she has x + 2 boxes, how many truffles does each box get, and will there be any left over?



Instruction

Resource List• Seward, Kim, for West Texas A&M University Virtual Math Lab. “Beginning Algebra Tutorial

30: Division of Polynomials.”

www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut30_divpoly.htm

This Web site offers a step-by-step tutorial and practice problems for long division of polynomials.

• Stapel, Elizabeth, for Purplemath. “Polynomial Division: Simplification and Reduction.”

www.purplemath.com/modules/polydiv2.htm

This site offers a step-by-step tutorial on long division of polynomials, delivered in a conversational tone, and accompanied by an interactive practice widget.





1. xx

− +−

411

4. xx

− ++

116

4

2. xx

+ ++

139

5 5. x

x− −

−7

713

3. xx

+ ++

1731


1. xx

− +−

1169

2. xx

+ −+

1821618

3. x xx

2 11 56104

4− + +

−

4. 11 16 29471

2x xx

− − −+

5. x xx

2 2 4123

6+ + +

−

6. x x xx

3 22 3 18424

+ − − ++

7. 2 2 8 50300

53 2x x x

x+ + + +

−

8. 5 15 35 4575

53 2x x x

x− + − +

+

9. x x xx

3 26 50120

6− − −

−

10. x x xx

3 210 15 2888

2+ + + +

−

11. 2 4488

xx

− +−

12. x xx

2 9 40150

10− + −

+

13. xx

+ −−

1750

6

14. xx

+ +−

1404

15. xx

+ ++

280

4

16. x xx

2 410

6+ + +

+


1. xx

− ++

3131

2. xx

+ +−

176210

3. x xx

2 28 61777

7− − −

−

4. 10 55 200745

52x x

x− + −

+

5. x xx

2 1124

6− + +

+

6. 3 27 162243

3x x

x3 + − +

+

7. 4 8 12 88452

43 2x x x

x+ + + +

−

8. x x xx

3 212 6 146108

3+ + + +

−

9. x x xx

3 24 45 2742029

7+ − + −

+

10. 4 3 8 1241

3 2x x xx

+ + − +−

11. xx

+ ++

119911

12. xx

− ++

683

13. x xx

2 10 100150

5+ − +

+

14. x2 – 5x + 12; 7 leftover tickets

15. xx

− ++

955

16. 2 5 424

22x x

x+ + −

−


1. x xx

2 3 480

4− + −

+ 4. x

x+ −

+16

42014

2. x xx

2 2 652

8+ − +

+ 5. x x

x2 5 46

2709

− + −+

3. xx

− −−

440

4 6. 7x2 + x + 7; 16 leftover truffles

Unit 2 • operations with algebraic expressionsLesson 11: Radical Expressions

naMe:


Assessment

Lesson Pre-AssessmentSimplify the following expressions.

1. 16 14 4 14 12 147 7 7y y y+ −

2. 250

6

63

3

x

3. 3 2 4 4 2 5b a b a+( ) +( )

4. The length of a rectangle is 24 y and the width is ( 150 ). What is the area of the rectangle?

5. What is the average speed in miles per hour of a car that travels 320x miles in 25 y hours?




Instruction

Lesson 11: Radical ExpressionsEssential Questions 1. How do you simplify algebraic expressions with radicals?

2. How is performing operations on expressions with radicals similar to performing operations on expressions with variables?

3. How do you apply arithmetic properties of radicals to rational expressions with radicals?

WORDS TO KNOW




index the small number n in the left part of the radical sign that becomes the exponent when rewriting a root xn as exponentiation


perfect square a number whose square root is a whole number

prime factor a prime number that evenly divides a number without any remainders

prime number a number divisible only by itself and 1


radicand the expression under the radical sign

radical sign a sign that indicates to take the root of a number

rationalizing rewriting a rational expression so that no radicals are in the denominator




naMe:


Warm-Up Option 1Find the prime factors of the following expressions.

1. 288

2. 64

3. 750

Write the following powers in standard form.

4. 53

5. 43

6. 36


naMe:



Warm-Up Option 2Use the given information to solve the following problems.

1. A box has a square base and a volume of 294 cubic inches.

a. Write an equation to represent each dimension of the box.

b. Calculate the actual dimensions of the box.

2. A house has the shape of a cube, and a volume of 216,000 cubic feet. What is the height of the house?



Instruction

Warm-Up Option 1: DebriefFind the prime factors of the following expressions.

1. 288

A prime number is a number that is divisible only by itself and 1.

• Examples include 2, 3, 5, 7, and 11.

Prime factors are prime numbers that divide a number evenly without a remainder.

• To find the prime factors of a number, divide repeatedly by prime numbers.

• Since 288 is even, it’s simplest to begin dividing it by 2.

288 ÷ 2 = 144

144 ÷ 2 = 72

72 ÷ 2 = 36

36 ÷ 2 = 18

18 ÷ 2 = 9

• When we can no longer divide by 2 and arrive at a whole number result, choose another prime number by which to divide; in this case, 3 is a logical choice.

9 ÷ 3 = 3

288 = 2 • 2 • 2 • 2 • 2 • 3 • 3

• This expression can be written more simply through the use of exponents.

2 • 2 • 2 • 2 • 2 • 3 • 3 = 25 • 32

2. 64

• To find the prime factors of the number, divide repeatedly by prime numbers.

64 ÷ 2 = 32

32 ÷ 2 = 16

16 ÷ 2 = 8

8 ÷ 2 = 4

4 ÷ 2 = 2

64 = 2 • 2 • 2 • 2 • 2 • 2 = 26




Instruction

3. 750

• To find the prime factors of the number, divide repeatedly by prime numbers.

• Use the rules of divisibility to determine the smallest prime number by which the number is divisible. In this case, we start out with 2, and then move on to 5.

750 ÷ 2 = 375

375 ÷ 5 = 75

75 ÷ 5 = 15

15 ÷ 5 = 3

• Simplify using exponents.

750 = 2 • 53 • 3

• Rearrange in order of least to greatest.

750 = 2 • 3 • 53

4. 54

• Express the exponentiation as repeated multiplication.

• Remind students that the exponent of 4 indicates the number of times the base of 5 is multiplied by itself.

• This type of problem is commonly misinterpreted as 5 • 4 = 20.

54 = 5 • 5 • 5 • 5 = 625

5. 43


43 = 4 • 4 • 4 = 64

6. 36


36 = 3 • 3 • 3 • 3 • 3 • 3 = 729



Instruction

Warm-Up Option 2: Debrief 1. A box has a square base and a volume of 294 cubic inches.

a. Write an equation to represent each dimension of the box.

• Refer to the volume formula for a prism.

Volume = length • width • height = l • w • h

• Identify the known information.

Volume = 294 cubic inches

• The base is a square; both the length and the width are equal.

• Assign one variable, x, to each.

length = x

width = x

• The height remains unknown. Assign it the variable h.

Volume = l • w • h

294 = x • x • h

294 = x2h

• To write an equation for the length, solve the equation for x.

294 = x2h

294 2

hx=

xh

=294




Instruction

• The length and width are equal, so the formula for width is also xh

=294

.

• To write the equation for height, solve the equation for h.

294 = x2h

294

2xh=

b. Calculate the actual dimensions of the box.

• The box has a square base. The length and width are the same, x, and it has a height, h. The volume is x2h.

• This means that we have to find two numbers such that the product of the square of the first number times the second number equals 294.

• Rewrite 294 as the product of its prime factors by dividing by prime numbers.

294 ÷ 2 = 147

147 ÷ 3 = 49

49 ÷ 7 = 7

294 = 2 • 3 • 7 • 7 = 2 • 3 • 7 2

• Use this expression to find the dimensions of the box.

Volume = x2 h294 = 7 2 2 • 3

x2 = 7 2, therefore x = 7

h = 2 • 3 = 6

The length and width must be 7 inches, and the height must be 6 inches.



Instruction

2. A house has the shape of a cube, and a volume of 216,000 cubic feet. What is the height of the house?

• If the shape is a cube, then length, width, and height are all equal.

• We have to find a number that, when cubed, equals 216,000.

• Rewrite the volume as a product of prime factors.

216,000 ÷ 2 = 108,000

108,000 ÷ 2 = 54,000

54,000 ÷ 2 = 27,000

27,000 ÷ 2 = 13,500

13,500 ÷ 2 = 6,750

6,750 ÷ 2 = 3,375

3,375 ÷ 5 = 675

675 ÷ 5 = 135

135 ÷ 5 = 27

27 ÷ 3 = 9

9 ÷ 3 = 3

• 216,000 = 26 • 33 • 53

• Find the combination of factors or the number that can be cubed to obtain 216,000.

216,000 = 26 • 33 • 53 = (22 • 3 • 5)3 = (4 • 3 • 5)3 = (60)3

Therefore, the height of the house is 60 feet.


naMe:



Focus ProblemNora and Sakura are training for a triathlon. Nora rides her bike a distance of 50x kilometers in

1812 hours and Sakura rides a distance of 512

12 x kilometers in 3 hours.

1. What is another way to write the expression of Nora’s time and Sakura’s distances with fractional exponents?

2. What is Nora’s speed in kilometers per hour?

3. What is Sakura’s speed in kilometers per hour?

4. Who is faster, and by how much?



Instruction


This problem has three parts. There are two people training for the triathlon, and both are riding different distances and times on their bikes.

• To calculate each person’s speed, divide their distance by their time.

• To calculate who is faster and by how much, subtract the speeds.

The example below will give you an idea of the process, but without radicals or variables, just pure arithmetic. You will see that though it may look complicated once you add variables and radicals, the process is very similar.

Example 1Zeph and Bea are training for the triathlon. Zeph rides her bike a distance of 30 kilometers in 3 hours and Bea rides a distance of 36 kilometers in 4 hours. Who is faster, and by how much?

Solution

• Calculate Zeph’s speed.

303

10= kilometers per hour

• Calculate Bea’s speed.

364

9= kilometers per hour

• Subtract the speeds to find who is faster and by how much.

Zeph – Bea = 10 – 9 = 1

• We subtracted Bea’s speed from Zeph’s and got a positive number because Zeph’s speed is the larger number of both speeds.

Zeph is faster by 1 kilometer per hour.




Instruction

Before continuing with the focus problem, use the following example as an introduction to simplifying radicals.

Example 2

Simplify the expression 83 .

Solution

• A root is the opposite operation of exponentiation.

If 32 = 9, then 9 3= .

If 23 = 8, then 8 23 = .

• The radical sign is the sign that we use to denote roots.

• The number under the radical sign that we are taking the root of is called the radicand.

• The small number in the left part of the radical sign is called the index, and it is comparable to the exponent in a power.

index

radical sign

8 23 =

radicand

• It is helpful to keep the above illustration posted while discussing radical expressions.

• When taking the square root, as in 9 , we do not need to write the index. It is implied that the degree is 2.

• For all other roots, we need to write the index, as in 83 .

• In the expression 83 , the index is 3 and the radicand is 8.

• This expression is read as “the cube root of 8.”

• To find the cube root of 8, find the factor of 8 that occurs 3 times.

2 • 2 • 2 = 23 = 8, so by definition, 8 23 = .



Instruction

Example 3

Simplify the expression 813 .

Solution

• Fractional exponents can be translated into roots. In general, x xn n1

= .

• Rewrite the given expression as a root.

125 12513 3=

• Rewrite the radicand, the expression under the radical sign, as a product of prime factors.

125 53 33=

• When the exponent of a factor in the radicand is equal to the degree of the radical sign, then that factor can leave the radical sign (the exponent and the degree cancel each other).

5 533 =

• The reason behind this is that having a radical sign is like having a fractional exponent.

5 5 5 5 533 313

313 1= ( ) = = =

( )•

125 513 =


Nora and Sakura are training for a triathlon. Nora rides her bike a distance of 50x

kilometers in 1812 hours and Sakura rides a distance of 512

12 x kilometers in 3 hours.





What is another way to write the expression of Nora’s time and Sakura’s distance with fractional exponents?

Instruction

• The expression for Nora’s time is 1812 .

18 1812 =

• Rewrite 18 as the square root of a product of primes.

18 2 3 3 2 3 3 22= • • = • =

• 2 32• can be rewritten as two separate radicals and then simplified.

2 3 2 3 3 22• = • =

Nora’s time is 18 3 212 = .

• The expression for Sakura’s distance is 51212 x .

• 51212 x can be thought of as two products, 512

12 and x.

512 51212

12x x= ( )

• Rewrite 512 as the square root of a product of primes.

x x x512 2 2 2 2 2 2 2 2 2 29= • • • • • • • • =• To simplify the radicand, look for powers that are multiples of the index.

• In this example, look for powers that are multiples of 2.

x x2 2 29 8= •



Instruction

• Rewrite x 2 28 • as separate products.

x x2 2 2 28 8• = •

• Simplify 28 .

x x x2 2 2 2 16 28 4• = • =

• Students having difficulty understanding this may benefit from seeing the radicand expanded.

x x2 2 2 2 2 29 2 2 2 2= • • • •

• x 2 2 2 2 22 2 2 2• • • • can be written as separate products.

x x2 2 2 2 2 2 2 2 2 22 2 2 2 2 2 2 2• • • • = • • • •

2 22 =

• Therefore, x x x2 2 2 2 2 2 2 2 2 2 16 22 2 2 2• • • • = • • • • = .

x x x x x x2 2 2 2 2 2 2 2 2 2 2 2 16 29 2 2 2 2 8 8 4= • • • • = • = • = =

• Another way to simplify the expression 51212 x is by using the properties of exponents.

• 512 can be rewritten as 29, as discovered earlier.

512 212 9

12x x= ( )

• 2912( ) can be rewritten as a product of powers.

2 2912

912( ) =

•




Instruction

• Simplify the power 912

• .

2 2 29

12

92

412

•= =

• Following the rules of exponents. the power of 412

can be rewritten as a sum of 4 and 12

.

2 2412

412=

+

• Rewrite 24

12

+ as 2 24

12• .

• 24 = 16 and 2 212 =

• Therefore, 512 16 212 x x= .

• Still another way to simplify the expression 51212 x is to look for perfect squares in the radicand:

512 512 2 256 2 16 16 212 2x x x x x= = • = • =

Therefore, Sakura’s distance is 512 16 212 x x= .

Question 2

What is Nora’s speed in kilometers per hour?

Instruction

• To find Nora’s speed, divide her distance in kilometers by her time in hours.

• Remember that in Question 1 it was determined that Nora’s time was 18 3 212 = .

• Nora’s speed = 50

3 2

x

• There is a radical in the denominator, so we need to rationalize the expression; in other words, we need to rewrite the fraction so that the denominator does not contain a radical.

• In order to rationalize the denominator, multiply the numerator and denominator of the fraction by the radical in the denominator.



Instruction

• This does not change the value of the fraction as it is similar to multiplying by 1.

50

3 2

50

3 2

2

2

50 2

3 2 2

x x x= • =

••

• Looking at the denominator, 2 2 2 22

• = ( ) = .

• Therefore, the expression 50 2

3 2 2

x ••

can be rewritten.

50 2

3 2 2

50 23 2

x x••

=••

• You can simplify the expression further by dividing 50 and 2 by a factor of 2.

50 23 2

25 23

x x••

=

Nora’s speed is 25 2

3x

kilometers per hour.

Question 3

What is Sakura’s speed in kilometers per hour?

Instruction

• To find Sakura’s speed, divide her distance in kilometers by her time in hours.

• Remember that in Question 1, we simplified Sakura’s distance from 51212 x to 16 2x .

• Her time is 3 hours.

• Sakura’s speed = 16 2

3x

• This expression is already simplified because the denominator is an integer and the radical is a prime number.

Sakura’s speed is 16 2

3x

kilometers per hour.





Who is faster, and by how much?

Instruction

• Both speeds are in simplified form.

• Pick an order to subtract the two speeds; for example, Nora’s speed – Sakura’s speed.

• If the difference is positive, it means Nora’s speed is the larger number and therefore she is the fastest. If the difference is negative, it means Sakura’s speed is the larger number and therefore she is the fastest.

25 23

16 23

x x−

• Both fractions have the same denominator, so we can subtract the numerators.

25 23

16 23

25 2 16 23

x x x x− =

−

• When subtracting the numerators, treat radicals as variables. If the radicals are similar, meaning they have the same index and radicand, they can be combined, just like variables.

• Factor the coefficients and treat the radical part as if it were a variable.

25 2 16 23

25 16 23

9 23

x x x x−=

−=

( )

• Be sure to look to reduce all fractions.

• Notice 9 and 3 have a common factor of 3.

9 23

3 2x

x=

The difference is 3 2x .



Instruction

• Is the difference positive or negative? It depends on the sign of the variable x. But we can assume x is positive: the scenario told us that “Nora rides her bike a distance of 50x kilometers.” Therefore, x must be positive; otherwise it would mean that Nora rode a negative distance, which is impossible!

• If x is positive, then the difference 3 2x is positive, too. That means that Nora is the fastest.

Nora is the fastest, by 3 2x kilometers per hour.


Simplify the expression 19 5 7 5− .

Solution

Adding or subtracting radical expressions is similar to adding monomials. If we were given instead the expression 19x – 7x, we could simplify it to 12x by subtracting the coefficients.

• Here we can do the same: Treat the radical as we would treat a variable, and subtract the coefficients.

19 5 7 5 19 7 5 12 5− = − =( )

19 5 7 5 19 7 5 12 5− = − =( )19 5 7 5 19 7 5 12 5− = − =( )

Example 2

Simplify the expression 75 6• .

Solution

The product of two radicals equals the radical of the product.

75 6 75 6 450• = • =

450 is not in the simplest form, since a number as large as 450 must have at least one perfect square as a factor.




Instruction

• Find factors of 450, of which at least one is a perfect square.

• For example, 9 and 50 are factors of 450.

• Remember that the radical of a product equals the product of radicals.

450 9 50 9 50 3 50= • = • =

• You can simplify this further by finding two factors of 50, of which at least one is a perfect square.

• 2 and 25 are factors of 50.

3 50 3 25 2 3 25 2 3 5 2 15 2= • = = • =

75 6 15 2• =

Example 3

Simplify the expression 63

7.

Solution

The quotient of two radicals is the equivalent to the radical of the quotient.

• Rewrite the expression, simplify the fraction, and then simplify the resulting radical if possible.

63

7

637

9 3= = =

63

7 = 3




Simplify the expression 2 7 3 2 5 3−( ) +( ) .

Solution

• Multiplying radical expressions in which some factors contain more than one term is analogous to multiplying polynomials.

• To simplify the expression above, apply the distributive property and FOIL.

2 7 3 2 5 3 2 2 2 5 3 7 3 2 7 3 5 3−( ) +( ) = • + • − • − •

• Multiply each term in the expression. Note that on the last term, 3 3 3 32• = = .

2 2 2 5 3 7 3 2 7 3 5 3 4 10 3 14 3 105• + • − • − • = + − −

• Simplify by combining like terms.

4 10 3 14 3 105 101 4 3+ − − = − −

Example 5

Simplify the expression 6

10.

Solution

To simplify an expression with a radical in the denominator, we must rationalize it, or rewrite the expression so that the denominator doesn’t have radicals.

• Multiply by a fraction with the same radical in both numerator and denominator.

• Here, we can multiply by 10

10, which equals 1 and therefore does not change the value

of the expression.

6

10

6

10

10

10

6 1010

3 105

= • = =

6

10

3 105

=


naMe:



Guided PracticeSimplify the following expressions.

1. 77 212x •

2. 4 3 12 75 2x x x+ −

3. ( )( )x x− +6 6

4. 32

4

2x

y

5. 6412

13

y

6. 2 45

37 53

x x−

7. 6

7

91

39 10

x

x x•

8. 2139

134x •

9. 72

4

18

36x x+

10. x5

23

16

continued


naMe:


Solve the following word problems.

11. Luke made 25x copies of a CD he recorded with his band. If he gives away 10, how many will he have left?

12. Wanda makes x 75 pounds of chili and serves it all in y plates with equal amounts. How many pounds of chili does each plate have?

13. A garden bed in the shape of a square has an area of 72y2 square feet. If the garden’s length is reduced by 5 feet and its width is increased by 5 feet, what is the new area of the garden bed?

14. Paulo is making balloon animals for a party. On Wednesday, he makes 196x balloon animals, and on Thursday he makes 400x . On Friday, the day of the party, he realizes that 225x balloon animals have popped. How many balloon animals does he have left?

15. Mason earns 330 dollars in 12 y hours. What is Mason’s pay per hour?

16. Jenna is trying to incorporate more spinach into her diet. She ate 308 y pounds of spinach over the course of 693 y days. How many pounds of spinach did she eat in all?


naMe:



Independent Practice Simplify the following expressions.

1. 12 15x x•

2. x x108 363+

3. x y10 8•

4. 100

100

x

y

5. 216 3

2

13y

x

6. x x806

203

−

7. 40

44

132

12 10

x

x x•

•

8. x 4

7

42

3•

9. 2 75

100

3 108

81

2y

x

y

x+

10. x

y

352

1249

( )( ) continued


naMe:


Solve the following word problems.

11. Cynthia has walked a total of 162 2x miles in a period of 3x days. If she walked the same amount each day, how far did she walk each day?

12. Ben made x 80 pounds of rice pudding and his family ate 5 2y pounds of it for dessert. How much rice pudding is left?

13. Jerome bakes 64 44 y cookies for a party, then bakes y 86 more. His brother eats 18 2y of the cookies. How many cookies are left?

14. Jade is a bread maker. She bakes 300b loaves of bread each week for 12b weeks. How many loaves of bread did she bake in total over that period?

15. Okkon has a rectangular garden bed with a length of 320x and a width of 245x. He wants to build another garden bed with the same area, but shaped as a square. What will be the length of the sides of the new square bed?

16. Beata has a square-shaped quilt with an area of 19a square feet. She cuts off 3a feet from the length, and then makes the quilt wider by attaching a new piece that is 3a wide and matches perfectly with the new length. What is the new area of her quilt?


naMe:



Assessment

Progress AssessmentSimplify the following expressions.

1. x x

y

54 250

128

3 3

63

+

2. x x

y

21

10

55

7

22

9 3

81

11• • •

3. Reilly is a member of two community gardens in the city. Every Saturday, he travels x 125

5

miles to the first garden, then y 52

13 miles to the second one, and finally

24

2 6

2z miles back

home. What’s the total distance he travels?

4. A sculpture has the shape of a cube and a volume of 512x3. What are the dimensions of the sculpture?

5. What is the average speed in miles per hour of a train that travels 21 45x miles in 49 y hours?

6. Louisa has a garden that is 8a yards wide and 27b yards long. She has decided that next year she will make it 3b yards wider and 2a yards longer. What will be the area of the garden bed after those changes?



Instruction

Resource List• Keeler, Alice, for Quia. “Radical Expressions.”


Simplify each radical expression to earn points. An incorrect answer results in the end of the game.

• Stapel, Elizabeth, for Purplemath.com. “Square Roots: Introduction and Simplification.”

www.purplemath.com/modules/radicals.htm

This site offers an introduction to square roots and operations with radicals, but no algebra. It has links to several sub-topics.





1. 8 147y 4. 60 y

2. 5 9

3

2 3x 5.

8

5

xy( )( )

3. 20 24 31 22a b a b+ +


1. 7 33x 9. 2 2x

x

2. x 3 10. x x3 3 28

3. x2 – 6 11. 5 2( )x −

4. 2 2x y

y 12.

5 3x y

y

5. 4

4y 13. 72y2 – 5

6. − x 53

14. 19 x

7. 5

5 15.

110

2

y

y

8. 3 72x 16. 462y


1. 6 5x 9. 3 3yx

2. 17 3x 10. x xy

y

7

7

3. 4 5x y 11. 3 2

4. 10x y

y 12. 4 5 5x y−

5. 6 3y xx

13. 0

6. 0 14. 60b

7. 1 15. 280x

8. x2 2 16. 16a


1. 2

2

xy

4. 8x

2. x y

y

2 11 5.

9 5x y

y

3. 5x + 2y + z 6. 4 9 7 6a b ab+ + cubic yards

Unit 2 • operations with algebraic expressionsMixed Review

naMe:


Solve. Show your work.

1. Evaluate the expression 3(8 – 6) + 4x – 5(z2 – 7) if x = 3 and z = 6.

Simplify the following expressions, noting the domain where necessary.

2. 7x5 – 2x + x5 + x2 – 4 + 8x

3. (22 + 5x2 – 4y) + (x2 + 8 + y) + (6x2 – 13y + 3)

4. (34z2 + 8z + 4) – (–3z2 – 9z + 26)

5. v8w5 • w4v • vw3 • v

6. (w4)5 • (v3)9 • (wv4)8

7. (5x3y7z9) ÷ (10w8x8y2z14)

8. a7b–5c8d–2 ÷ a3b4c–6d–3

9. (x2 + 12x + 27) ÷ (x + 9)

10. − + −

−x x x

x

12 12

144

3 2

2

continued


naMe:



Simplify the following radical expressions.

11. + −x x x5 3 20 45 2 15. +x x

50

3

8

49

12. − +x x( 7)( 7) 16.

y

16

72

3

13. x

y

50

100 2 17. Reduce

−− +x

x x

3 108

12 36

2

2 to the simplest form.

14. •x1530

32

Multiply the following polynomials.

18. (6m – 4n)2 19. (a2 – b)(4a + 9b + 8ab)

Factor the following polynomials.

20. 25 – g2 21. z2 + 12z + 35

continued


naMe:


Complete the following problems by first writing the appropriate polynomial expression.

22. Cell phone Company A charges $0.30 per minute with no flat fee per month. Cell phone Company B charges $0.10 per minute with a flat fee of $25 per month. Write an expression and determine the cost of each company’s cell phone plan if 175 minutes are used in one month.

23. The flight of a bird can be represented by a polynomial with respect to the ground. One bird flies according to the polynomial (–4r2 – 3r + 9). Another bird flies according to the polynomial (2r2 + 12r + 6). What is the sum of the birds’ flight paths?

24. A projectile is shot from point P, a distance described by the polynomial 8t2 + 4t + 3. The projectile is then shot from the point where it landed back to point P, but this time it only travels the distance described by the polynomial 5t2 + 6t. How far from the initial point P is the projectile?

25. Bradley is building a shed for his father. He decides on the width and length, which will give the shed an area of t2 + 8t + 16. What might be the length and width of the sides?

26. A city block has an area of x2 + 12x + 70 square meters. If the city block is x + 9 meters wide, how long is it?



Unit 2 • OperatiOns with algebraic expressiOns

Mixed Review Answer Key 1. –127

2. 8x5+x2+6x–4

3. 12x2–16y+33

4. 37z2+17z–22

5. v11w12

6. v59w28

7.y

w x z2

5

8 5 5

8.a c d

b

4 14

9

9. x+3

10.+

+x

x

1

12

2

;domain:allrealnumbersexceptx=–12,12

11. x4 5

12. x2–7

13.x

y

2

2

14. x5 3

15.x

x

41 2

21

16.y y2

8

4 23

17.+−

x

x

3( 6)

( 6),x≠6

18. 36m2–48mn+16n2

19. 4a3+9a2b+8a3b–4ab–9b2 – 8ab2

20. (5–g)(5+g)

21. (z+5)(z+7)

22. 0.30x;0.10x+25;52.50;42.50

23. –2r2+9r+15

24. 3t2–2t+3

25. (t+4)and(t+4)

26. + ++

xx

343

9

Unit 2 • operations with algebraic expressionsUnit Post-Assessment 1

naMe:


Assessment


1. 3a2 – 4b, where a = 3, b = 5

2. 26x + 5y – 2x – 8xy – 9y + 7xy

3. (5a2 + 10a) + (6a2 + 11)

4. (y2 – 6y + 15) – (3 + 2y – 4y2)

5. (m3n)2 • (3m2)4 • mn5

continued


naMe:



Assessment

6. a –5b –6d 7 ÷ a –7b 7c –4d –6

7. (2x – 3y) (4x2 – xy + 4y2)

8. (16 + k) (16 – k)

9. Factor 16x2 – 4b2.

10. 72 42 54

6

5 4 3

2

x x xx

− +−

continued


naMe:


Assessment

11. x x

x x

2

2

4 3220 96

− −− −

12. (x3 + 13x2 + x – 50) ÷ (x + 3)

13. (3x3 – 12x2 + 6x – 48) ÷ (x – 4)

14. x x

y

16 25

18

+

15. x 18

8 +

y 150

6 –

36

4 9

2x




Unit Post-Assessment 1 Answer Key 1. 7 2. 24x – 4y – xy 3. 11a2 + 10a + 11 4. 5y2 – 8y + 12 5. 81m15n7

6. a dc b

2 13

4 13

7. 6x3 – 14x2y + 11xy2 – 12y3

8. 256 – k2

9. (4)(2x – b) (2x + b) 10. –12x3 + 7x2 – 9x

11. ( )( )

xx

−−

824

, x ≠ –4, 24

12. x xx

2 10 2937

3+ − +

+ 13. 3 6

244

2xx

+ −−

14. x

y3

2

15. x + 5y


naMe:


Assessment


1. 5a3 – 6b, where a = 2, b = 4

2. 34xy – 6y + 3x + 7xy + 8y – 5xy

3. (12a2 + 15a) + (4a2 + 8)

4. (3y2 + 8y + 9) – (6 – 7y – y2)

5. (m4n)3 • (2m3)2 • mn6

continued


naMe:



Assessment

6. a–3b–4d7 ÷ a–9b5c–2d–4

7. (5x – 4y)(3x2 – xy + 7y2)

8. (11 – k)(11 + k)

9. Factor 125x2 – 5y2.

10. + −−

x x x

x

32 72 16

8

6 5 4

3

continued


naMe:


Assessment

11. − −− −

x x

x x

3 18

6 27

2

2

12. (x3 + 17x2 + 25x – 161) ÷ (x + 5)

13. (5x3 – 30x2 + 15x – 112) ÷ (x – 6)

14. +x x

y

49 36

8

15. x 162

3 2 +

x64

2 16

2

– y 45

5




Unit Post-Assessment 2 Answer Key 1. 16

2. 3x – 2y + 36xy

3. 16a2 + 15a + 8

4. 4y2 + 15y + 3

5. 4m19n9

6. a c d

b

6 2 11

9

7. 15x3 – 17x2y + 39xy2 – 28y3

8. 121 – k2

9. (5)(5x – y)(5x + y)

10. –4x3 – 9x2 + 2x

11. −−

x

x

( 6)

( 9) , x ≠ –3, 9

12. + − ++

x xx

12 3514

52

13. + −−

xx

5 1522

62

14. x

y

13 2

4

15. 4x – 3y

Unit 2 • operations with algebraic expressionsStation Activities Set 1: Evaluating and Simplifying Expressions


InstructionGoal: To provide opportunities for students to develop concepts and skills related to evaluating

and simplifying expressions

Student Activities Overview and Answer KeyStation 1

Students roll a number cube to get values for three variables. Then they work together to evaluate a set of expressions that contain these variables. Students are encouraged to make sure everyone agrees on each answer before writing it down, and they are also asked to reflect on the steps they used to evaluate the expressions.

Answers: Answers will vary depending upon the numbers that are rolled.

Possible steps: In each expression, replace each variable by its value. Then simplify the numerical expression using the order of operations and the rules of integer arithmetic.

Station 2

Students are given a set of cards with algebraic expressions written on them. Students work together to sort the cards into pairs so that the cards in each pair show equivalent expressions. Then they explain the strategies they used to solve the problem.

Answers: The cards should be paired as follows: 3p + 4n + 3p and n + 6p + 3n, n + 3p and 8p + n – 5p, 4n + 3p – 2n and 2n + 3p, 2n + 8p – 3n and 2p + 6p – n, 2(3p – n) and 10p – 2n – 4p

Possible strategies: First simplify the expressions on the cards with expressions that can be simplified. Then look for another card that contains the simplified expression or an expression that can be simplified to the same expression.

Station 3

Students are given a set of cards with simple algebraic expressions on them. Students choose four cards at random and write the expressions in boxes provided on the activity sheet. In this way, students generate expressions that they will simplify by working together.

Answers: Answers will vary depending upon the cards that are chosen. Note that it is possible to generate expressions that cannot be simplified.




Instruction

Station 4

Students work together to match a set of given expressions with a set of integer values of the variable. The goal is to pair expressions and values of the variable so that every expression has a value of 24 when evaluated for the value of the variable with which it is paired. All students should agree on the pairing of the cards before writing the answer.

Answers: The cards should be paired as follows: 2(x + 2) and x = 10, 6x 2 and x = –2, 2x 2 + 6 and x = –3, 6(5 + x) and x = –1, 30 – 3x and x = 2.

Possible strategies: Choose an expression and evaluate it for each possible value of the variable until the result is 24. Alternatively, choose an expression and decide which value of the variable makes it equal 24, and then match it with this value.

Station 5

The class will form teams and use blocks to represent algebraic expressions. Then teams will combine expressions and find the sum of their expressions. This station uses concrete manipulatives to help students understand the abstract concept of combining similar terms.

Answers: Round 1: Teams 1, 2 = , Teams 3, 4 =, and Teams 5, 6 = ; Round 2: Teams 1, 4 = , Teams 2, 5 = , and Teams 3, 6 = ; Round 3: Teams 1, 5 = , Teams 2, 3 = , and Teams 4, 6 =

Materials List/SetupStation 1 number cube (numbers 1-6)

Station 2 set of 10 index cards with the following expressions written on them: 3p + 4n + 3p, n + 3p, 4n + 3p – 2n, n + 6p + 3n, 2n + 8p – 3n, 2(3p – n), 2n + 3p, 10p – 2n – 4p, 8p + n – 5p

Station 3 set of 10 index cards with the following expressions written on them: 2x, 4x, 2x 2, 3x 2, 6x 2, 3y, 4y, 2y 2, 3y 2, 8y 2

Station 4 set of 5 index cards with the following expressions written on them: 2(x + 2), 6x 2, 2x 2 + 6, 6(5 + x), 30 – 3x set of 5 index cards with the following values of x written on them: x = –3, x = –2, x = –1, x = 2, x = 10

Station 5 6 containers with 11 interconnecting blocks of one color and 8 interconnecting blocks of another color

6 6 2 2 22 2 2y b by y b+ + + + 6 4 3 32 2y by y b+ + +3 3 22 2 2 2y by y b y b b+ + + + 6 4 4 32 2y b y b+ + +

6 2 2 22 2 2 2y b by y b b+ + + + 3 7 22 2 2y by y b y b+ + + +6 4 2 22 2 2 2y b y y b b+ + + + 6 2 4 2 22 2 2y b by by y b+ + + + +

6 2 4 2 22 2 2y b by by y b+ + + + + 3 3 2 32 2 2y by y y b b+ + + +



Instruction

Discussion GuideTo support students in reflecting on the activities and to gather some formative information about student learning, use the following prompts to facilitate a class discussion to “debrief” the station activities.

Prompts/Questions

1. How do you evaluate an expression for a given value of the variable or variables?

2. After you substitute the value of the variable or variables in the expression, how do you simplify the result?

3. How do you evaluate the expression 12n 2 + t for specific values of n and t?

4. What steps do you use to simplify an algebraic expression?

5. How do you determine that terms are like terms?

Think, Pair, Share

Have students jot down their own responses to questions, then discuss with a partner (who was not in their station group), and then discuss as a whole class.

Suggested Appropriate Responses

1. Substitute the value of the variable(s) for the variable(s) in the expression and simplify.

2. Perform the operations, using the order of operations and the rules of integer arithmetic.

3. Substitute the values for n and t in the expression. Square the value of n by multiplying it by itself. Then multiply by 12 and add the value of t.

4. Look for like terms. Combine the like terms by adding or subtracting coefficients.

5. Like terms have the same variables and the same power on those variables.

Possible Misunderstandings/Mistakes

• Applying an incorrect operation (e.g., adding instead of multiplying when evaluating for a value of x)

• Incorrectly applying the order of operations when evaluating or simplifying an expression that involves more than one operation

• Incorrectly applying the distributive property (e.g., writing 3(x + 1) = 3x + 1)

• Incorrectly constructing blocks

• Assuming that by b y2 2=


naMe:



Station 1You will find a number cube at this station.

Roll the number cube three times. Write the numbers in the boxes below. This will give you values for the variables m, q, and s. (Note that the value of s is a negative number.)

Work with other students to use these values to evaluate each expression. When everyone agrees on an answer, write it on the line.

1. 4m + q ________ 5. 3s 2 ________

2. 3q + s ________ 6. s – m ________

3. 4qs ________ 7. 60m

+ q ________

4. –2mq + 3 ________ 8. 5(q – m) ________

Explain the steps you used to evaluate the expressions.

m = q = s = –


naMe:


Station 2You will find a set of ten cards at this station. The cards contain the following expressions:

3p + 4n + 3p n + 3p 4n + 3p – 2n n + 6p + 3n

2n + 8p – 3n 2(3p – n) 2n + 3p 10p – 2n – 4p

8p + n – 5p 2p + 6p – n

Work together to sort the cards into pairs. The cards in each pair should show equivalent expressions. When everyone agrees on the answer, write the five pairs below.

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

Explain the strategies you used to solve this problem.


naMe:



Station 3You will find a set of cards at this station. The cards should be spread out, face-down.

Choose four cards without looking. Write the expressions on the cards in the boxes below.

Work together to simplify the expression. When everyone agrees on the answer, write it below.

Simplified expression: ________

Put the cards back. Mix up the cards. Then repeat the above process four more times.





+ + +

+ + +

+ + –

+ – –

+ – +


naMe:


Station 4At this station, you will find five cards with the following expressions written on them:

You will also find five cards with the following values of x written on them:

Work together to match each expression with a value of x so that when you evaluate each expression, the result is 24.

Work together to check that each pair gives a value of 24. When everyone agrees on the results, write the five pairs below.

_____________________________

_____________________________

_____________________________

_____________________________

_____________________________

Describe the strategies you used to solve this problem.

2(x + 2) 6x 2 2x 2 + 6 6(5 + x) 30 – 3x

x = –3 x = –2 x = –1 x = 2 x = 10


naMe:



Station 5At this station, you will be combining similar or like terms. You will find a container with 11 blocks of one color and 8 blocks of another color. You will construct a model of terms that corresponds to your team number. See the example below for help in constructing your model.

Form the construction that corresponds to your team number.

Team 1:

Team 2:

Team 3:

Team 4:

Team 5:

Team 6:

Check your construction with your teacher.

Team # ________

Algebraic Term Block Construction Picture2b 2 single second color

2by 2 constructions, each having 1 block of 1st color and 1 block of 2nd color

2y 2 2 squares of blocks in 1st color

b 2y 1 square of blocks 2nd color connected with one yellow

3 4 2 2 2y b y b+ + +3 2 2 2 2y b by b+ + +3 4 2 2y by y b+ + +3 2 22 2y y b+ +3 2 2 2y y b+3 2 2by y b b+ +

continued


naMe:


Teams will now pair up, combine their blocks, and write down the algebraic expression that represents their result.

Round 1

Team combinations: Teams 1 and 2, 3 and 4, 5 and 6

Your team’s algebraic expression: _____________________

Team # ___________’s algebraic expression: _____________________

The algebraic expression of the total: _____________________

Round 2





Round 3





Unit 2 • operations with algebraic expressionsStation Activities Set 2: Operations with Polynomials



InstructionGoal: To provide opportunities for students to develop concepts and skills related to adding,

subtracting, multiplying, and dividing polynomials


Students will be given 20 blue algebra tiles, 20 red algebra tiles, 20 green algebra tiles, and 20 yellow algebra tiles. Students work together to model polynomials with algebra tiles. Then they add the polynomials using the algebra tiles.

Answers

1. 8 52 2x xy y+ +

2. Answers will vary. Possible answer: We combined same-color algebra tiles.

3. Answers will vary. Possible answer: We used the zero property to find pairs of same-color algebra tiles that canceled each other out.

4. zero property

5. 12y2 – 12xy – 5x2 – 4

6. Answers will vary.

7. Answers will vary. Possible answer: We used the zero property.

8. 5a3 – 3a2b2 + 10b3

9. 10x2 – y2 – 15xy + 4

10. 16c3 – 8a3 + 3ac2 – 7

Station 2

Students will be given 20 blue algebra tiles, 20 red algebra tiles, 20 green algebra tiles, and 20 yellow algebra tiles. Students will work together to model polynomials with algebra tiles. Then they will subtract polynomials using the algebra tiles.

Answers

1. 5x2 + 5xy + 4y2


3. 3x2, 2xy, 2y2

4. –5x2 – 5xy – 4y2



Instruction

5. No, because the sign of the terms in the second polynomial changes to the opposite sign.

6. 7x2 + 5xy + 11y2

7. Answers will vary. Possible answer: We matched like terms, and then performed subtraction.

8. Answers will vary. Possible answer: We added negative terms because subtracting a negative number is the same as adding a positive of that number.

9. –2a4 – 4a2b2 + 6b3 + 6

10. 6c2 – 6bc – 18

Station 3

Students will be given a number cube. Students will use the number cube to populate coefficients of polynomials. Then they will multiply polynomials using the distributive property.

Answers

1. Answers will vary. Possible answer: 2x and (3x + y – 2)

2. distributive property

3. Answers will vary. Possible answer: 2 3 2

6 2 42

x x y

x xy x

( )+ −

+ −

4. Answers will vary. Possible answer: –3x2 and (–4x + 7xy – 8)

5. Answers will vary. Possible answer: 12x3 – 21x3y + 24x2

6. It changed to the opposite sign because we multiplied each term by –1.

7. (x + 3) and (x – 4)

8. distributive property

9.

( )( )x x

x x x

x x

+ −

− + −

− −

3 4

4 3 12

12

2

2

10. We combined –4x and 3x.

Station 4

Students will be given six index cards with the following polynomials written on them:

3xy2; 18x2 – 7x + 4; 33xy5 – 3x2y2 – 21xy2; 2x; –3y2; –24y5 + 6y3 – 12




Instruction

Students will work together to match polynomials and monomials that when divided by each other yield a specific quotient. Then students will perform synthetic division to divide a polynomial by a binomial.

Answers

1. 18 7 4

29

72

22x xx

xx

− += − +

2. − + −

−= − +

24 6 12

38 2

45 3

23

2

y yy

y yy

3. 33 3 21

311 7

5 2 2 2

23xy x y xy

xyy x

− −= − −

4. Answers will vary. Possible answer: We divided each term by the monomial using the quotient rule for exponents.

5. No, because you have to divide by a binomial instead of a monomial.

6. 3; 3

7. Find the solution of the binomial, which is 1. Write this in the left-hand box. Write the coefficients of the variables in a row. Bring down the first coefficient. Multiply this coefficient by 1. Add this product to the second coefficient. Repeat this process through all the coefficients. The last number is the remainder.

8. 2 6 741

3 2x x xx

+ + +−

9. 4 5 15 3157

23 2x x x

x+ + + +

−

Materials List/Setup

Station 1 20 blue algebra tiles, 20 red algebra tiles, 20 green algebra tiles, and 20 yellow algebra tiles

Station 2 20 blue algebra tiles, 20 red algebra tiles, 20 green algebra tiles, and 20 yellow algebra tiles

Station 3 number cube

Station 4 six index cards with the following polynomials written on them:

3xy2; 18x2 – 7x + 4; 33xy5 – 3x2y2 – 21xy2; 2x; –3y2; –24y5 + 6y3 – 12



Instruction


Prompts/Questions

1. How do you add polynomials?

2. How do you subtract polynomials?

3. What happens to the exponents of the variables when you add or subtract polynomials?

4. How do you multiply polynomials?

5. How do you deal with the exponents of the variables when multiplying polynomials?

6. How do you divide a polynomial by a monomial?

7. How do you divide a polynomial by a binomial?

8. How do you deal with the exponents of the variables when dividing polynomials?

Think, Pair, Share



1. Add like terms.

2. Subtract like terms of the second polynomial from like terms in the first polynomial.

3. Exponents remain the same in addition and subtraction of polynomials.

4. Multiply each term in the first polynomial by each term in the second polynomial using the distributive property.

5. Use the product rule on the exponents.

6. Divide each term in the polynomial by the monomial.

7. Use synthetic division.

8. Use the quotient rule on the exponents.




Instruction


• Incorrectly adding exponents when adding polynomials

• Incorrectly subtracting exponents when subtracting polynomials

• Not using the product rule on exponents when multiplying polynomials

• Not using the quotient rule on exponents when dividing polynomials

• Not using synthetic division when dividing by a binomial

• Not realizing that the last number in synthetic division is the remainder


naMe:


Station 1At this station, you will find 20 blue algebra tiles, 20 red algebra tiles, 20 green algebra tiles, and 20 yellow algebra tiles. Work as a group to model each polynomial by placing the tiles next to the polynomials. Then find the sum.

• Use the blue algebra tiles to model the x2 term.

• Use the red algebra tiles to represent the xy term.

• Use the green algebra tiles to represent the y2 term.

• Use the yellow algebra tiles to represent the constant.

1. Given: 3 2 2

5 3

2 2

2 2

x xy y

x xy y

+ +

+ − ++

3 2 2

5 3

2 2

2 2

x xy y

x xy y

+ +

+ − +

Answer: __________________

2. How did you use the algebra tiles to model the problem?

3. How did you model the –xy term?

4. What property did you use on the xy terms? __________________

5. Model the following problem using the algebra tiles. Show your work.

( ) ( )4 12 5 10 8 42 2 2 2y xy x x y− + + − + −

Answer: __________________ continued


naMe:



6. How did you use the algebra tiles to model problem 5?

7. How did you deal with negative terms during addition?

Work together to add each polynomial. Show your work.

8. Given: 2a3 + a2b2 + 3b3

+ 3a3 – 4a2b2 + 7b3

9. –10xy – 3 + 2x2 – 5y2 + 4y2 + 8x2 – 5xy + 7

10. 8c3 + 3ac2 + 4a3 + 8c3 – 12a3 – 7


naMe:


Station 2At this station, you will find 20 blue algebra tiles, 20 red algebra tiles, 20 green algebra tiles, and 20 yellow algebra tiles. Work as a group to model each polynomial by placing the tiles next to the polynomials. Then find the difference.

• Use the blue algebra tiles to model the x2 term.

• Use the red algebra tiles to represent the xy term.

• Use the green algebra tiles to represent the y2 term.

• Use the yellow algebra tiles to represent the constant.

1. Given: 8 7 6

3 2 2

2 2

2 2

x xy y

x xy y

+ +

− + +( )–

8 7 6

3 2 2

2 2

2 2

x xy y

x xy y

+ +

− + +( )

Answer: __________________

2. How did you use the algebra tiles to model the problem?

3. What terms in the bottom polynomial does the subtraction sign apply to?

4. Find the difference: 3x2 + 2xy + 2y2

– (8x2 + 7xy + 6y2)

Answer: __________________

5. Is your answer from problem 1 the same as your answer from problem 4? Why or why not?

continued


naMe:



6. Model the subtraction problem below using the algebra tiles, then solve. Show your work.

2x2 + 5y2 + 9xy

– (4xy – 5x2 – 6y2)

Answer: __________________

7. How did you arrange the algebra tiles to model problem 6?

8. How did you deal with negative terms during subtraction?

Work together to subtract each polynomial. Show your work.

9.

a a b b

a a b b

4 2 2 3

4 2 2 3

4 8

3 3 2 2

− + +

− + − +( )–

a a b b

a a b b

4 2 2 3

4 2 2 3

4 8

3 3 2 2

− + +

− + − +( )

10. Subtract 8c2 + 2bc + 10 from –4bc + 14c2 – 8.


naMe:


Station 3At this station, you will find a number cube. As a group, roll the number cube. Write the result in the empty box below for the given equation.

Given: x x y( )3 2+ −

1. Identify the two polynomials above: __________________

2. What property can you use to multiply these polynomials? __________________

3. Multiply the polynomials. Show your work.

Roll the number cube again. Write your result in the empty box below.

Given: − − + −x x xy2 4 7 8( )



continued


naMe:



6. What happened to the signs of each term of the polynomial in the parentheses? Explain your answer.

Given: (x + 3)(x – 4)


8. What method can you use to multiply these polynomials? __________________


For problem 10, fill in the empty box with the number you rolled for problems 4 and 5 on the previous page.

10. What extra steps did you take when multiplying (x + 3)(x – 4) versus − − + −x x xy2 4 7 8( ) ?


naMe:


Station 4At this station, you will find six index cards with the following polynomials written on them:

3xy2; 18x2 – 7x + 4; 33xy5 – 3x2y2 – 21xy2; 2x; –3y2; –24y5 + 6y3 – 12

Shuffle the cards. Work as a group to match the polynomials that when divided yield each quotient below. (Hint: Place the monomials in the denominator.)

1. 972

2x

x− +

2. 8 2432y y

y− +

3. 11y3 – x – 7

4. What strategy did you use for problems 1–3?

Given: ( ) ( )2 4 3 13 2x x x x+ + − ÷ −

5. Can you use the same strategy to divide the polynomials above as you did in problems 1–3? Why or why not?

6. What is the degree of ( )2 4 33 2x x x+ + − ? __________________ This degree means the

quotient will have __________________ terms.

continued


naMe:



Follow these steps to use synthetic division to find the quotient of ( ) ( )2 4 3 13 2x x x x+ + − ÷ − .

Step 1: Set the binomial equal to zero and solve for x. Show your work.

Step 2: Use your answer from Step 1 and write it in the first box on the left in the illustration below.

2 6 7

2 6 7 4−

Step 3: Write the coefficients of each term in order of left to right in the top row of boxes in the illustration under Step 2.

Step 4: The first coefficient (in this case, 2 of 2 • 3) is always written in the box underneath it.

Step 5: The boxes that are already filled in show synthetic division.

7. Derive the process of synthetic division based on the example above.

Step 6: The last number in the bottom row, –4, is known as the remainder and can be written

as 41x −

.

8. Based on steps 1–6, what is the answer to ( ) ( )2 4 3 13 2x x x x+ + − ÷ − using synthetic division? __________________

9. Use synthetic division to find ( ) ( )4 3 5 5 24 3 2x x x x x− + + − ÷ − . Show your work.

Unit 2 • operations with algebraic expressionsStation Activities Set 3: Factoring Polynomials


InstructionGoal: To provide opportunities for students to develop concepts and skills related to factoring

polynomials


Students will be given a number cube. Students will use the number cube to populate the exponents of terms and expressions. They will find the greatest common factor of terms and expressions. Then they will factor the expression using the greatest common factor.

Answers

1. Answers will vary. Possible answer: x3, x6, x4; x3

2. 1; x3; x

3. greatest common factor

4. Answers will vary. Possible answer: 4x3 – 6x2 + 4x4; 2; x2; 2x2; 2x – 3 + 2x2

5. Answers will vary. Possible answer: –5x2y + x3 – 10x5y4; 1; x2; x2; –5y + x – 10x3y4

6. Answers will vary. Possible answer: 6x3yz + 2x + 4x2y5; 2; no common variables; 2; 3x3yz + x + 2x2y5

7. No, because there was no variable that all three terms had in common.

8. 4x2y4z3(3x2y + 14z3 – 6xy3z5)

9. 3c2(9a2b3 – 4ac – 3b2c3)

10. –9s2t(4r2st + 2rs + 3r2s2t 4 + 1)

Station 2

Students will be given eight blank index cards, plus 10 index cards with the following written on them:

3x, x, +1, +2, +4, +8, –1, –2, –4, –8

Students will work together to arrange the cards to factor a trinomial. Then they will create the possible factors of a trinomial and factor the trinomial. Students factor trinomials with a leading coefficient other than 1.

Answers

1. (3x + 2)(x + 4)

2. Answers will vary. Possible answer: We used the distribution method to check factors.




Instruction

3. The factors are 3x and x because 3 is a prime number.

4. x, 2x, 3x, and 6x

5. –5, –1, 1, 5

6. (2x – 1)(3x + 5)

7. Use the distribution method to double-check answers.

Station 3

Students will be given five index cards with the following expressions written on them:

x x2 8 12+ + ; x x2 8 15− + ; x x2 2 80+ − ; x x2 12+ − ; x x2 12− −

They will also receive five index cards with the following factors written on them:

( )( )x x− +3 4 ; ( )( )x x+ −10 8 ; ( )( )x x+ +2 6 ; ( )( )x x+ −3 4 ; ( )( )x x− −3 5

Students will work together to match each expression with the appropriate factors. Then students will factor trinomials with a leading coefficient of 1. They will explain how to double check their answers and why factoring out the greatest common factor first is important.

Answers

1. x x2 8 12+ + and ( )( )x x+ +2 6

2. x x2 8 15− + and ( )( )x x− −3 5

3. x x2 2 80+ − and ( )( )x x+ −10 8

4. x x2 12+ − and ( )( )x x− +3 4

5. x x2 12− − and ( )( )x x+ −3 4


7. Use the distribution method to multiply the binomials. This should yield the original trinomial.

8. 1

9. x and x

10. 6 and –2; ( x + 6 )( x + –2 ); (x + 6)(x – 2)

11. 2

12. x2 + 4x – 5



Instruction

13. x and x

14. –5; 4; 5 and –1; ( x + 5 )( x – 1 ); 2(x + 5)(x – 1)

15. Answers will vary. Possible answer: It’s easier to factor smaller numbers.

Station 4

Students will be given a number cube. Students will use the number cube to populate binomial expressions. They will multiply the binomial expressions using the distribution method. Then they will factor the polynomial they created. They will relate the distribution method to factoring. They will factor the difference of squares and perfect square trinomials.

Answers

1. Answers will vary. Possible answer: (x + 2)(x – 2) = x2 – 4

2. 2

3. It cancels out.

4. (2x + 3)(2x – 3); Find the square root of the first term and third term. Write the factors in (a + b)(a – b) form.

5. (7x3 + 6)(7x3 – 6)

6. Answers will vary. Possible answer: (2x + 3)(2x + 3) = 4x2 + 12x + 9

7. 3

8. (4x + 3)(4x + 3); Find the square root of the first term and third term. Write the factors in (a + b)(a + b) form.

9. (2x4 + 5)(2x4 + 5)

10. Answers will vary. Possible answer: (x – 3)(x – 3) = x2 – 6x + 9

11. 3

12. (5x – 3)(5x – 3); Find the square root of the first term and third term. Write the factors in (a – b)(a – b) form.

13. (6x2 – 1)(6x2 – 1)




Instruction

Materials List/SetupStation 1 number cube

Station 2 eight blank index cards; 10 index cards with the following written on them:

3x, x, +1, +2, +4, +8, –1, –2, –4, –8

Station 3 five index cards with the following expressions written on them:

x x2 8 12+ + ; x x2 8 15− + ; x x2 2 80+ − ; x x2 12+ − ; x x2 12− −

five index cards with the following factors written on them:

( )( )x x− +3 4 ; ( )( )x x+ −10 8 ; ( )( )x x+ +2 6 ; ( )( )x x+ −3 4 ; ( )( )x x− −3 5

Station 4 number cube



Instruction


Prompts/Questions

1. How do you find the greatest common factor of terms with variables?

2. How do you factor a trinomial with a leading coefficient not equal to 1?

3. How do you factor a trinomial with a leading coefficient equal to 1?

4. How do you factor the difference of two squares?

5. How do you factor the perfect square trinomial a2 + 2ab + b2?

6. How do you factor the perfect square trinomial a2 – 2ab + b2?

Think, Pair, Share



1. Find the greatest common factors of the coefficients. Find the variable with the lowest exponent that can be divided into each term of the polynomial.

2. Find the factors of the leading coefficient. Find the factors of the last term that add up to the middle term taking into account the factors of the leading coefficient.

3. Find the factors of the last term that add up to the middle term taking into account x and x as the first terms. (Assuming the first term is x2.)

4. Take the square root of the first term and the third term. Put it in the form (a – b)(a + b).

5. Take the square root of the first term and third term. Put it in the form (a + b)(a + b).

6. Take the square root of the first term and third term. Put it in the form (a – b)(a – b).




Instruction


• Not factoring out the greatest common factor first

• Not using law of exponents correctly when factoring

• Not finding the factors of the third term that add up to the middle term when factoring trinomials

• Not canceling out the middle term when factoring the difference of two squares


naMe:


Station 1At this station, you will find a number cube. As a group, roll the number cube. Write your answer in the box below. Repeat this process until all the boxes contain a number. The numbers will be the exponents for the variables.

x , x , x

1. Of the three terms above, which term has the lowest exponent? __________________

2. Divide your answer from problem 1 into each of the three terms above. Write your answers below.

__________________

__________________

__________________

3. You found the largest monomial that could be divided into all the terms. What is the name for this factor? __________________

Roll the number cube to populate the boxes for each problem below.

4. 4x – 6x + 4x

What is the greatest common factor of the coefficients? ______

What is the greatest common factor of the variables? ______

What is the greatest common factor of the three terms? ______

Factor out the greatest common factor of each term. Show your work.

continued


naMe:



5. –5x y + x – 10x y





6. 6x yz + 2s + 4x y





7. Did you factor out any variables in problem 6? Why or why not?

continued


naMe:


8. What is the greatest common factor of the following equation?

12x4y5z3 + 56x2y4z6 – 24x3y7z8

______


27a2b3c2 – 12ac3 – 9b2c5

______


–36r 2s3t 2 – 18rs3t – 27r 2s4t 5 – 9s2t

______


naMe:



Station 2At this station, you will find eight blank index cards, plus 10 index cards with the following written on them:

3x, x, +1, +2, +4, +8, –1, –2, –4, –8

As a group, determine which index cards to use to factor:

3 14 82x x+ +

1. What are the factors of 3 14 82x x+ + ? ________________

2. How did you determine which index cards to use in problem 1?

3. Why were 3x and x the only factors of 3x2 ?

Given: 6 7 52x x+ −

4. What are the factors of 6x2? __________________

Write each factor on separate index cards.

5. What are the factors of –5? __________________

Write each factor on separate index cards.

6. As group, arrange the index cards you created to help you factor 6 7 52x x+ − .

7. How can you double-check to see if you factored the trinomial correctly?


naMe:


Station 3At this station, you will find five index cards with the following expressions written on them:

x x2 8 12+ + x x2 8 15− + x x2 2 80+ − x x2 12+ − x x2 12− −

You will also find five index cards with the following factors written on them:

( )( )x x− +3 4 ( )( )x x+ −10 8 ( )( )x x+ +2 6 ( )( )x x+ −3 4 ( )( )x x− −3 5

Shuffle the cards. As a group, match the expressions with their factors. Write the matches on the lines below.

1. __________________

2. __________________

3. __________________

4. __________________

5. __________________

6. What strategy did you use to match the cards?

7. How can you double-check your matches?

continued


naMe:



Given: x x2 4 12+ −

8. What is the greatest common factor of all three terms? ______

Use your answers for problems 9 and 10 to fill in the boxes below.

( + )( + )

9. What are the factors of x2? __________________

Write these factors in the solid boxes.

10. What are the factors of –12 that add up to 4? __________________

Write these factors in the dashed boxes.

The factors of x x2 4 12+ − are __________________.

continued


naMe:


Given: 2 8 102x x+ −

11. What is the greatest common factor of all three terms? ______

12. Factor out the greatest common factor. What is the new expression? __________________

Use your answers for problems 13 and 14 to fill in the boxes below.

( + )( + )

13. What are the factors of x2? __________________

Write these factors in the solid boxes.

14. What are the factors of ______that add up to ______?

Write these factors in the dashed boxes.

The three factors of 2x2 + 8x – 10 are __________________

15. Why should you factor out the greatest common factor first before factoring the expression?


naMe:



Station 4At this station, you will find a number cube. As a group, roll the number cube twice. Write the first number that you roll in the first empty box below. Write the second number you roll in the second empty box.

( x + )( x – )

1. Use the distribution method to multiply the two binomials. Show your work.

2. How many terms does the polynomial you created in problem 1 contain?

__________________

3. Why is there no x term in the polynomial you created in problem 1?

4. How can you factor 4x2 – 9 using the observations you made in problems 1–3? Show your work.

5. Factor 49x6 – 36. Show your work.

continued


naMe:


As a group, roll the number cube twice. Write the first number that you roll in the first empty box below. Write the second number you roll in the second empty box.

( x + 3 )( x + 3 )



__________________

8. How can you factor 16x2 + 24x + 9 using the observations you made in problems 6 and 7? Show your work.

9. Factor 4x8 + 20x4 + 25. Show your work.

continued


naMe:



As a group, roll the number cube once. Write this number in both boxes below.

( x – )( x – )



__________________

12. How can you factor 25x2 – 30x + 9 using the observations you made in problems 10 and 11? Show your work and answer.

13. Factor 36x4 – 12x2 + 1. Show your work.

Unit 2 • operations with algebraic expressionsStation Activities Set 4: Simplifying and Operations with Square Roots


InstructionGoal: To provide opportunities for students to develop concepts and skills related to square

roots, particularly in context of simplifying, adding, subtracting, multiplying, and dividing


Students use a number cube to create radicals with two-digit radicands. Students work together to decide if each radical shows a square root that can be simplified. If so, students work together to write the simplified square root. Students make sure that each person in the group agrees with the answer.

Answers: Answers will vary depending upon the numbers rolled.

Station 2

Students are given ten numbers or expressions that involve radicals. The numbers and expressions are written on index cards and students work together to find pairs of cards that represent equal numbers. To do so, students need to simplify, add, and subtract square roots. Students write equality statements based on the pairs of matching cards.

Answers: 24 2 6 75 2 3 3 3 32 4 2 8 5 2 3 2 20 2 5

2 8 18 32

= = + = = − =; ; ; ;

, , , ,, , , ,

, , , , , , ,

, ,

50 72 98 128

2 2 2 3 2 5 2 2 3 4 3 5 3 7 3

24 4 2 755 2 6 5 2 3 2 2 5 8 2 3 3 3 20 32, , , , , , ,− +

Possible strategies: First simplify the square roots that can be simplified. Find the sum or difference of the square roots on cards that involve addition or subtraction.

Station 3

Each student writes the square root of a prime number (such as 3

1

4

5 5 5

6 12 6 2 5 20 10 2 1032

24

24

32 2

2

× =

× = × = × = =; ; ; ;

, 66 2 2 10 3 2 3 5 15

6 12 10 2 6 12524

32 2 2 1

, , , , , ,

, , , , ,× × × 00 6 2 102 4 5 2032

2

15 5 3 32 16 2

5 5 25

, , , ,× ×

= =

× =

) on a slip of paper. The slips of paper are placed in a container and students choose two slips of paper without looking. Students use these two square roots to create an expression involving division of square roots. Students work together to simplify the expression by rationalizing the denominator. Then they repeat the process several times by choosing new square roots.

Answers: The answers will depend upon the square roots that are chosen.

Station 4

Students are given cards with square roots written on them. They use the cards and a penny to create various expressions involving multiplication and division of square roots. Students work together to perform the required operation and simplify the resulting expression.

Answers: Answers will depend upon the numbers that are chosen.




Instruction

Materials List/SetupStation 1 number cube (numbered 1–6)

Station 2 10 index cards with the following numbers or expressions written on them:

24 2 6 75 2 3 3 3 32 4 2 8 5 2 3 2 20 2 5

2 8 18 32

= = + = = − =; ; ; ;

, , , ,, , , ,

, , , , , , ,

, ,

50 72 98 128

2 2 2 3 2 5 2 2 3 4 3 5 3 7 3

24 4 2 755 2 6 5 2 3 2 2 5 8 2 3 3 3 20 32, , , , , , ,− +

Station 3 blank slips of paper small empty container (for example, a plastic food container, an empty soup can, or a coffee mug)

Station 4 penny 8 index cards with the following numbers written on them:

3

1

4

5 5 5

6 12 6 2 5 20 10 2 1032

24

24

32 2

2

× =

× = × = × = =; ; ; ;

, 66 2 2 10 3 2 3 5 15

6 12 10 2 6 12524

32 2 2 1

, , , , , ,

, , , , ,× × × 00 6 2 102 4 5 2032

2

15 5 3 32 16 2

5 5 25

, , , ,× ×

= =

× =



Instruction


Prompts/Questions

1. How do you simplify a square root?

2. When is it possible to simplify an expression by adding or subtracting square roots? Give a specific example.

3. Give an example of an expression involving a sum or difference of square roots that cannot be simplified.

4. How can you use a calculator to check that you have correctly simplified an expression involving square roots?

5. How do you multiply two square roots?

6. How do you divide two square roots?

Think, Pair, Share



1. Write the radicand as a number times a perfect square. The square root of the perfect square can be simplified to a whole number.

2. You can add or subtract when the expression involves square roots of the same number. (For example, 3 2 4 2 7 2

3 5

15 5 3

32 16 2

2 3 5

3 3

+ =

+

=

=

+ =

− = −

)

3. The expression

3 2 4 2 7 2

3 5

15 5 3

32 16 2

2 3 5

3 3

+ =

+

=

=

+ =

− = −

cannot be simplified.

4. Use the calculator to evaluate the original expression and the simplified expression. The two numerical values should be the same.

5. Write the product of the radicands under a radical. Then simplify the resulting square root.

6. Write the quotient of the radicands under a radical. Then simplify the resulting square root. Alternatively, rationalize the denominator and then simplify.




Instruction


• Incorrectly simplifying square roots (e.g., writing

3 2 4 2 7 2

3 5

15 5 3

32 16 2

2 3 5

3 3

+ =

+

=

=

+ =

− = −

or

3 2 4 2 7 2

3 5

15 5 3

32 16 2

2 3 5

3 3

+ =

+

=

=

+ =

− = −

)

• Incorrectly combining square roots that cannot be added or subtracted (e.g., writing

3 2 4 2 7 2

3 5

15 5 3

32 16 2

2 3 5

3 3

+ =

+

=

=

+ =

− = −

)

• Distributing a minus sign across a radical (e.g., writing

3 2 4 2 7 2

3 5

15 5 3

32 16 2

2 3 5

3 3

+ =

+

=

=

+ =

− = − )

• Incorrectly multiplying identical square roots (e.g., writing

3

1

4

5 5 5

6 12 6 2 5 20 10 2 1032

24

24

32 2

2

× =

× = × = × = =; ; ; ;

, 66 2 2 10 3 2 3 5 15

6 12 10 2 6 12524

32 2 2 1

, , , , , ,

, , , , ,× × × 00 6 2 102 4 5 2032

2

15 5 3 32 16 2

5 5 25

, , , ,× ×

= =

× = )

• Failing to rationalize a denominator


naMe:


Station 1You will find a number cube at this station. You will use the number cube to create square roots.

Roll the number cube two times. Write the two numbers in the boxes inside the radical sign below.

Work with other students to decide if the square root can be simplified. If so, write the simplified square root below. If not, write “cannot be simplified.” Be sure everyone in the group agrees with the answer.

___________________

Repeat the process four more times.

___________________

___________________

___________________

___________________

0000000000

0000000000

0000000000

0000000000

0000000000


naMe:



Station 2You will be given a set of cards with the following numbers or expressions written on them:

Work with other students to find pairs of cards that show the same number. When you have paired the cards, work as a group to check that the numbers in each pair are equal.

Write five statements that use an equal sign (=) to list the pairs of equal numbers.

___________________

___________________

___________________

___________________

___________________

Write at least two strategies you could use to help you decide which numbers were equal.

24 4 2 75 2 6 5 2 3 2

2 5 8 2 3 3 3 20 32

−

+

24 4 2 75 2 6 5 2 3 2

2 5 8 2 3 3 3 20 32

−

+

24 4 2 75 2 6 5 2 3 2

2 5 8 2 3 3 3 20 32

−

+

24 4 2 75 2 6 5 2 3 2

2 5 8 2 3 3 3 20 32

−

+

24 4 2 75 2 6 5 2 3 2

2 5 8 2 3 3 3 20 32

−

+

24 4 2 75 2 6 5 2 3 2

2 5 8 2 3 3 3 20 32

−

+

24 4 2 75 2 6 5 2 3 2

2 5 8 2 3 3 3 20 32

−

+

24 4 2 75 2 6 5 2 3 2

2 5 8 2 3 3 3 20 32

−

+

24 4 2 75 2 6 5 2 3 2

2 5 8 2 3 3 3 20 32

−

+

24 4 2 75 2 6 5 2 3 2

2 5 8 2 3 3 3 20 32

−

+


naMe:


Station 3At this station, you will find some blank slips of paper and an empty container. Each student should take one slip of paper.

Write the square root of a prime number on your slip of paper (for example, or ).

Place all the slips of paper in an empty container.

Choose two slips of paper without looking. Write the square roots that you chose in the boxes below to create a quotient of square roots.

Work together to simplify the expression by rationalizing the denominator. When everyone agrees on the answer, write it below.

___________________


7 11


naMe:



Station 4You will find a set of cards and a penny at this station. Use these to create multiplication and division problems involving square roots.

Choose two cards without looking. Turn the cards over. Write the numbers in the two boxes below.

Flip the penny. If the penny lands heads up, write “” on the line below. If it lands tails up, write “” on the line.

Work with other students to find the product or quotient of the numbers. Simplify the answer if possible. When everyone agrees on the answer, write it below.

___________________


___________________

___________________

___________________

___________________



Unit Glossaryadditive inverse the original term but with the opposite sign, such that when added to of a monomial the original term the sum equals zero

additive inverse the original polynomial with each of its terms replaced with their of a polynomial additive inverses




commutative property in a multiplication problem, the product remains the same even of multiplication if the order of the factors is changed

constant a quantity that does not change


dividend in a division problem, the number that is the whole divided in parts

divisor in a division problem, the number that divides the dividend

domain the values that x can take within a certain rational expression



expression a symbol or combination of symbols representing a value or relation


greatest common in algebra, the greatest monomial that is a factor of all the terms in a factor (GCF) polynomial or algebraic expression

index the small number n in the left part of the radical sign that becomes the exponent when rewriting a root xn as exponentiation



negative exponent an exponent with a negative sign in front of it; indicates how many times to divide by a number

naMe:

Unit 2 • operations with algebraic expressionsUnit Glossary

naMe:



order of operations the order in which expressions are evaluated from left to right (parentheses, exponents, division and multiplication, and addition and subtraction—PEMDAS)

perfect square a number whose square root is a whole number



prime factor a prime number that evenly divides a number without any remainders

prime number a number divisible only by itself and 1

quotient the result of division

radical sign a sign that indicates to take the root of a number

radicand the expression under the radical sign


rationalizing rewriting a rational expression so that no radicals are in the denominator

remainder in a division problem, the portion of the dividend that does not divide exactly into the divisor, and that is left after dividing

simplest form a rational expression whose common factors between the numerator and denominator have been canceled and cannot be simplified any further




acr ea unit 2 - apple

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