algebra & geometry

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By Brad Fulton Educator of the Year, 2005

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c + f + 2s =


Strategies for Success: Algebra and G

eometry

Brad Fulton TTT Press

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By Brad Fulton Educator of the Year, 2005

b r a d @ t t t p r e s s . c o m w w w . t t t p r e s s . c o m 5 3 0 - 5 4 7 - 4 6 8 7

P.O. Box 233, Millville, CA 96062

c + f + 2s =


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Known throughout the country for motivating and engaging teachers and students, Brad has co-authored over a dozen books that provide easy-to-teach yet mathematically rich activities for busy teachers while teaching full time for over 30 years. In addition, he has co-authored over 40 teacher training manuals full of activities and ideas that help teachers who believe mathematics must be both meaningful and powerful.

Seminar leader and trainer of mathematics teachers Ø 2005 California League of Middle Schools Educator of the Year Ø California Math Council and NCTM national featured presenter Ø Lead trainer for summer teacher training institutes Ø Trainer/consultant for district, county, regional, and national workshops

Author and co-author of mathematics curriculum Ø Simply Great Math Activities series: six books covering all major strands Ø Angle On Geometry Program: over 400 pages of research-based geometry instruction Ø Math Discoveries series: bringing math alive for students in middle schools Ø Teacher training seminar materials handbooks for elementary, middle, and secondary school

Available for workshops, keynote addresses, and conferences All workshops provide participants with complete, ready-to-use activities that require minimal preparation and give clear and specific directions. Participants also receive journal prompts, homework suggestions, and ideas for extensions and assessment. Brad's math activities are the best I've seen in 38 years of teaching! Wayne Dequer, 7th grade math teacher, Arcadia, CA “I can't begin to tell you how much you have inspired me!” Sue Bonesteel, Math Dept. Chair, Phoenix, AZ “Your entire audience was fully involved in math!! When they chatted, they chatted math. Real thinking!” Brenda McGaffigan, principal, Santa Ana, CA “Absolutely engaging. I can teach algebra to second graders!” Lisa Fellers, teacher

References available upon request

BradFultonE d u c a t o r o f t h e Y e a r

¨ Consultant ¨ Educator ¨ Author ¨ Keynote presenter ¨ Teacher trainer ¨ Conference speaker

PO Box 233, Millville, CA 96062 (530) 547-4687

b r a d @ t t t p r e s s . c o m


Table of Contents Using Language to Develop Number Sense ................................................. 7

Number sense can be fostered using simple, language-based strategies. These techniques can be used at any time and make great opening math talks. The principles of using written and oral language can be applied to any mathematical topic in addition to developing number sense.

Pyramid Math............................................................................................. 22 This patterned approach to working with numbers helps students develop number sense while providing engaging practice. This will work with whole numbers, decimals, integers, and algebraic expressions.

Menu Math ................................................................................................. 32 This seven-course meal provides all the nutrition necessary for students young and old to grow strong and healthy algebraic brains. This sensible approach helps all students prepare for algebra by taming concepts such as variable, substitution, algebraic properties, equation solving, and systems of equations.

Acquisition of Geometric Understanding ................................................... 59 The learning of geometry has been stratified into five levels. While most students are operating at the lowest level, high school geometry requires them to work at the highest levels. Fortunately this problem can be remedied through these simple steps.

Graphic Organizers for Teaching Algebra .................................................. 69 Many students don’t know where to begin when tackling an algebra problem. These classroom-tested graphic organizers helped my students make significant gains on their state test scores and alleviated much of the, “I don’t get it.” Learn how to help students with binomial multiplication, factoring polynomials, combined work problems, and percent mixture problems.

“X” Marks the Spot .................................................................................... 73 This is also a great way to foster number sense in addition, subtraction, multiplication, and division of whole numbers, fractions, decimals, and integers.. The simple format will also transition students seamlessly into much higher mathematics.

Multiplying and Factoring Binomials ......................................................... 89 Using the graphic organizers students have seen makes quick work of even the most challenging polynomials.

Notes, Grids, and Evaluation .................................................................... 104


The Number Sense Problem As we improve our performance on high-stakes tests, number sense skills seem to be declining. This may be due to more procedural instruction that favors better performance on state tests. Number sense is a top priority in lower grades, but it is often replaced with more abstract instruction as children get older. Like physical health, our “number health” atrophies with a lack of exercise. Research shows that number sense is developed only through frequent instruction over a period of years. We cannot cover number sense as a separate unit of instruction as we can with other areas of mathematics. It cannot be learned from a chapter in a textbook nor assessed with standardized testing. Yet we know when a student has it and when he does not. Like great writing, number sense is hard to define yet immediately recognizable. Though it eludes definition, we can recognize it by its characteristics. Students with number sense have an ease and fluidity with numbers and their operations. They demonstrate coordination with mathematics much like a gymnast shows coordination in physical movement. Students with number sense can take numbers apart and recompose them in more expedient ways. This is what allows a student to see that 9+7=10+6 or that 24x3=8x3x3=8x9. A sense about numbers also allows a student to know when an answer is reasonable. Such students have not only a sense for the magnitude of numbers, but also of the meaning of a mathematical problem. This helps them know if their answer fits the situation or if an error is present. At its core, number sense is built upon a solid foundation in place value. It has been said that if you have place value, you have number sense, and if you have number sense, you have place value. Yet this is only the foundation and not the entire structure. Much of higher mathematics shows a separation from place value. Exponents are one example of this. Though 99 is composed of single digits, that is those in the ones place, it is an extremely large number. Algebra, geometry, and other branches of higher mathematics include many areas that are far beyond the scope of simple place value. As students get older, we must be sure we are preparing them with a solid foundation in “algebra sense”, “geometry sense”, “trigonometry sense” and so on. Fortunately, not only can we identify the characteristics of number sense but also the strategies that foster its development. As shown in the following pages of this handout, the inclusion of simple components of classroom instruction will help your students to show great growth in their number sense skills.


The Five Components Of Number Sense

Ø Estimation Ø Mental Computation Ø Mathematical properties Ø Effects of operations Ø Number magnitude


Estimation

Estimate these problems: 1. Which is greater, or ? Why? 2. 3. 4. 5. 6. 7. Write as a percent.

86×38 88×36

0.52×789

40×26.7

43×52

4953÷68

0.4×58.6×5×3

7151866


Answers 1. 86x38=3,268 while 88x36=3,168. Two sets of numbers have the same sum, the pair

with the lesser difference will produce a greater product. 2. 410.28 3. 1,068 4. 2,236 5. Approximately 72.8 6. 351.6 7. Approximately 38.3%


stimation relies on two fundamental skills: approximation and mental computation. According to Case, the

typical child is not capable of handling both of these tasks until around age 11. R. Case Intellectual Development 1985

E


esearch has shown that…“For students to become highly skilled at estimation, it had to be incorporated into their regular

instruction over several years.” From Research Ideas for the Classroom: Middle Grades Mathematics Douglas T. Owens

R


Mental Computation

Solve these problems mentally if possible.

1.

2. 45% of 250

3. 3467–1588

4. Find a fraction between and .

5. Find a fraction between and .

6.

16×15

1118

57

57

1318

5×2,379


Answers: 1. 240 2. 112.5 3. 1,879 4. 2/3 is one example 5. 18/25 is one example 6. 11,895


Mathematical properties

Use mathematical properties to solve each problem. Explain how the properties work.

1.

2. 37+256+19+50

3. 9(4,555)

4. 10,002–4,566

5. 18÷1.5

6. 1.5÷

16×52

23


Answers: 1. 832. The distributive property can be applied. You could also cut 16 in half while

doubling 52. Eventually you will get 1x832. This is an application of the associative property of multiplication.

2. 362. The associative property allows us to rearrange the addends this way: 37+19+(44+6)+256=

(37+19+44)+(6+256)= 100+262=

362 3. 40,995. The distributive property can be used two ways:

9(4,555)= 9(4,000+500+50+5)=

36,000+4,500+450+45 or

9(4,555)= (10–1)(4,555)= 45,000–4,555=

45,000–4,000–500–5-–5 4. 5,436. 10,002–4,566=9,999–4,563. This is called the compensation method of

subtraction.

5. 12. 18÷1.5=36÷3. If we think of the division problem as a fraction: , then we

have multiplied it by one in the form of .

6. 2.25. Using the method above, we can multiply by to get . Half of that

would be 2.25.

181.5

22

1.523

33

4.52


Effects of operations

Explain your answer to each question and give examples when necessary.

1. If two numbers are added, will the sum always be greater?

2. If you subtract one number from another will the result be smaller than the first number?

3. If numbers are multiplied is the product always greater?

4. When you square a number, is the answer always greater?

5. If divide one number by another, will the quotient always be smaller?

6. If you keep adding one to both the numerator and denominator of , how big will the number eventually be?

12


Answers: 1. No, if you add zero or a negative number, the sum is not greater. 2. No, if you subtract zero or a negative number, the difference is greater. 3. If you multiply by one or less, the product is equal or less. 4. The square will be greater unless the value of the number is between zero and

one. 5. If you divide by a number x such that 0<x≤1, the quotient of not greater. 6. Eventually the fraction approaches 1.


Number magnitude

1. How long will it take to live a million

seconds?

2. How long will it take to live a billion seconds?

3. How long will it take to live a trillion seconds?

4. How tall are a million dollar bills?

5. If you cut your paper in half twenty times, how big would it be?

6. How many times will you blink in your life?


Answers: 1. About 11.6 days 2. 31.7 years 3. 31,700 years 4. Over 650 feet tall. Each dollar is 1/125” thick. 5. Less than one ten-thousandth of a square inch, or about 1/100” by 1/100” 6. Between 300 billion and 500 billion times. Estimates vary.


Strategies that Develop Number Sense

Ø Playing with numbers Ø Solving problems in multiple ways Ø Creative practice Ø Thinking, talking, and writing about

numbers Ø Exploring patterns


Pyramid Math Building numbers sense is of critical importance if students are to be successful now and in later mathematics courses. But how do we develop number sense and provide skills practice at the same time? Fortunately we are providing a simple format for accomplishing both of these goals. This activity can be tailored to suit any grade level, and it can even be made self-correcting. Students will prefer this practice method to the standard textbook assignment, and they will be able to measure their own success. We even provide five student activity masters and answer keys! Procedure: 1 Display the Activity Master 4 and enter four numbers

in the cells of the bottom row as shown. (Use single-digit whole numbers at first so students can focus on the structure of the problem instead of stumbling on the computation at this point.) To solve the pyramid, an adjacent number pair is added. The sum is written in the box above the number pair. This is repeated for the other number pairs in the bottom row. Then this process is repeated for the second row to fill the third row. Finally the number pair in the third row is added to get the final top number as shown.

2 Since each sum is based on the sums below, all students should get the same

answer in the top cell. Thus they only need to check the top answer. If that is correct, all other cells are likely correct too.

3 Now try another pyramid using new numbers. Students will catch on to the process

quickly and will be eager to check their answer with those of their classmates. (No more correcting papers!)

4 As students understand how the problems work, introduce appropriate

numbers. If you are studying decimals, throw in a few decimal points. If you have covered integers, use some negative numbers. Fractions make these problems much more difficult. Try one yourself before asking the students to do so. We suggest beginning with like denominators. Or you could use fractions that have a fairly small common denominator. For example, ¼, ½, ½, and ¾ can all use fourths for a common denominator.

5 Try to make slight variations in the arrangement and values of the numbers to help children

focus on the number sense involved. For example, in the first problem if we increase the five

Required Materials: ý Paper ý Activity Master Optional Materials: ¨ Transparency of Activity

Master

“Building Mathematical Skill on a Foundation of Understanding”

Volume 303 March, 2003

5 2 1 8

5 2 1 8

7 3 9

10 12

22


by one, making it a six, the top number also increases by one, but if we change the two to a three, the top number increases by three.

Why does this occur? What would happen if we added two to the first or second cell of the bottom row? What happens when we do this to a five-row pyramid? As students answer these questions they will develop number sense.

6 Ask students to change the order of the numbers in the bottom row of a pyramid.

How does this affect the top cell? Is the result in the top row always the same? How does the commutative property affect this result?

7 Introduce subtraction as shown in this sample. You can create one of your

own easily, or have students create them for their classmates to solve. 8 Explore what happens when all odd numbers or all even numbers are used.

What if all four cells in the bottom row contain the same number? What patterns occur?

9 The pyramid at the right shows the algebra involved in these problems.

Notice that the numbers in the two middle cells get added three times on the way to the top while the numbers in the two outer cells do not. This explains why increasing the first cell by one causes the top row to increase by one, but increasing either of the two middle cells by one increases the top row by three.

We always love to hear from our readers. Email us with your

comments on this activity at either [email protected] or [email protected]

6 2 1 8

8 3 9

11 12

23

5 3 1 8

8 4 9

12 13

25

7

17

21

52

a b c d

a+b b+c c+d

a+2b+c b+2c+d

a+3b+3c+d

Good Tip! C To make the activity self-assessing, write the answers to the problems at the bottom of the student’s worksheet. As a student solves a problem, the answer can be crossed off from the list. If the answer is not found, the problem can be solved again.


5

4 5

3 2

1

12

6

9 5

2

7

13

31 7

3

19

15

18 11

4

12

4

5 15

5

73

27

56 44

6

105 61

38 22

7 53

33

21

13

8 84

56

21

14

9

Activity Master 1 Pyramid Math

Name______________________________

Add adjacent numbers and write their sum in the box above them as shown in the example. Continue until your reach the top of the pyramid. On some problems, you may need to work backwards.


.05

4 2.3

.3 2

1

1.2

.06

.99 .5

2

.7

13

3.1 .07

3

.19

15

1.8 1.1

4

1.2

4.6

.05 1.5

5

.73

.27

5.6 4.4

6

14 4.3

.50 .5

7 6

3.3

2.0

1.3

8 8.4

6.4

3

1.5

9


Name______________________________



-5

4 -1

-3 2

1

5

-4

-3 2

2

-5

4

3 -2

3

19

-15

18 -11

4

12

-4

-5 15

5

73

27

-56 -44

6

-3 4

7 24

7 17

33

12

13

8 13

-21

-21

14

9


Name______________________________



1

2

3

4

5

6

7

8

9


Name______________________________

Add adjacent numbers and write their sum in the box above them. Continue until your reach the top of the pyramid. On some problems, you may need to work backwards.


1

2

3

4

5

6


Name______________________________

Add adjacent numbers and write their sum in the box above them. Continue until your reach the top of the pyramid. On some problems, you may need to work backwards.

7

8


Answer Key Answers are provided for the top box of the pyramid except for problem 7 of each set. The answer for the left box of the bottom row is provided for problem 7. Activity Master: Problem: 1 2 3 4 5 6 7 8 9 1 28 62 146 119 54 366 14 94 140 2 14.05 4.85 49.07 51.69 16.65 62.34 2.1 10.7 14.8 3 0 -14 14 18 0 -58 -18 13 -8

Research Says: Though many researchers have trouble explaining exactly what number sense is or how to develop it, most of us as teachers can recognize when a student does or does not have number sense. By looking at a wide variety of research samples however, a framework of number sense components starts to take shape. Five of the most important components are 1) Estimation; 2) Mental computation; 3) Mathematical properties; 4) Mathematical operations; and 5) Number magnitude. In addition, numerous strategies begin to emerge for fostering number sense in students. They are 1) Toying with numbers; 2) Multiple computational methods; 3) Creative practice; 4) Exploring patterns; and 5) Use of written and oral language. Providing students with opportunities to use these strategies will help them build the components of number sense. Keep in mind though that textbooks don’t tend to utilize these strategies. That is because the purpose of a text is to increase math skill, not number sense. However, both are necessary. This means that as teachers we will need to work beyond the book to build number sense. The development of number sense is an ongoing endeavor that results more from experiences by the student than from direct instruction. Activities like Pyramid Math can be used to provide these experiences. Further Help in Developing Number Sense! For more activities that provide students with opportunities to develop number sense, take a look at our recently revised book on number sense. The second edition includes even more pages and a more teacher-friendly format that makes your job much easier. Students will enjoy making sense of numbers while building their computational skills.

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Characteristics of Number Sense

Ø Nonalgorithmic

Ø Tends to be complex

Ø Involves meaning

Ø Relies on judgment

Ø Is “effortful”

Ø Relies on self-regulation of the thinking process

Ø May involve uncertainty

Ø May yield multiple solutions or multiple solution paths


umber sense is something that “unfolds” rather than something that is “taught” directly.

P. R. Trafton Establishing foundations for research on number sense and related topics 1989

N


PROCEDURE Skills:

ý Using variables

ý Solving equations

ý Problem solving

ý Distributive Property

1. Display the transparency master on the overhead projector as the students get out paper. Use a paper to cover the formulas, revealing only the menu at this time.

2. Slide the paper down to reveal the first formula:

h + f = ? Students will raise their hands to tell the answer. One will say,

“Two dollars and ninety cents!” Ask the student, “What do you mean?” The response will be that a burger and fry cost two dollars and ninety cents. Ask how the student arrived at that answer. Everyone will think it is obvious that the “h” stands for hamburger and the “f” must represent fries.

3. Display the second formula:

c + f + s = ? Again a student will volunteer to explain that a cheeseburger,

fries, and small soda costs $4.15. You may wish to ask how the student decided that the “s” must stand for “small” and not for “soda”. Again, the class will see that this is the obvious way to interpret the formula since there are three sizes of sodas.

4. Reveal the third formula:

7f = ? Students will say that the cost is now $7.35. Ask the class how

they got that answer. Someone will say that you need to multiply seven times $1.05. Ask why the student decided to multiply when the formula had no multiplication sign in it? Most students will think the answer is obvious. No other operation would make sense.

Menu Math Overview: This activity is so easy to teach you might feel guilty for taking a paycheck! Students will quickly pick up on the meaning and use of variables and will be able to substitute values into formulas to solve them. Vocabulary: variable, formula

Materials: ý paper o transparency of activity

master

ACTIVITY 4


5. At this stage, students can solve the next two formulas for further practice.

6. The final formula requires students to solve for an unknown,

which is letter “d” in this case. Once they realize that the “d” is equal to $1.55, they should see that it represents a large soda. You can then ask what the “d” might stand for. Someone may suggest it stands for “drink”.

7. At this point, continue to ask the students to solve more

complex formulas and equations depending on their ability. Some samples are given here:

3c + 5f + 6s = 6f + 4s = 8h + 4f + 6x = 11c + 8m + 2x = 5c + 2h + m + 2s = 3h + 2s + m = 3h + s = 2c + d (What does d = ?)

Jason gave the clerk a ten dollar bill and received $3.95 in change. What might he have ordered?

× × × × Ø Ø Ø Ø

Journal Prompts: ? Here is a formula for an order at a restaurant. Write the dialog the

customer may have used when placing the order. 2h + c + 3f + 2x + m

Homework: & Ask students to make up formulas and solve them. These

formulas can be created toward the end of class and then copied down by the students to be solved for homework.

Taking a Closer Look: If you had ten dollars, what different combinations of food could

you order? Try to spend as much of the money on each order without exceeding $10.00. Write each answer as a equation. For example:

Good Tip! C

Students can create their own menus and formulas and exchange them with a partner.


3h + 3f + m = $9.95 Ask the students to solve this formula:

4h + 4s + 4x = This formula can be used to illustrate the distributive property

by solving 4(h + s + x) = and seeing that the same answer results. You can then have the students practice the distributive property by solving similar formulas.

Assessment: þ Students can verify each other's formulas if working in groups.

Homework can be checked through a class discussion.


Menu Math

Hamburger .................................... $1.85 Cheeseburger ................................ $2.15 Fries .............................................. $1.05 Sodas: Small .......................................... $.95 Medium .................................... $1.25 Large ........................................ $1.55 Extra Large .............................. $2.05 h + f =

c + f + s =

7f =

3h + c + f + 3x =

4c + 3f + s + m + l =

3c + 3d = $11.10 What does d = ?


Write what each customer ordered and calculate how much was paid for each order: 3h + 3f = 3h + f = 3(h + f) = Which two customers ordered the same food and paid the same price? Write the two orders below:

__________ = __________

This is called the "Distributive Property."


Write the following orders as algebraic expressions.

I'd like four hamburgers, six orders of French fries, a large soda, two medium sodas, and an extra large soda.

I want three cheeseburgers, one hamburger,

a small soda, two fries, a medium soda, and another hamburger.

I want a cheeseburger and an order of fries

with a medium soda, my son wants two hamburgers an order of fries, and a medium soda, and my daughter wants a cheeseburger, an order of fries and a large soda. Oh yes, my husband wants two orders of fries, a cheeseburger and a large soda.

Let's see… I think I'd like three hamburgers and a cheeseburger, three fries, a large soda, two medium sodas, and an extra large soda. Add another order of fries on that, and make one of those hamburgers another cheeseburger.


Write a dialog or skit for each of the following algebraic orders.

3x + h + c 2h + 2f + 2m (h + 2f) + (2c + x) (3h + f + x) + (h + f + s) (h + f + m) + (h + f + m) + (h + f + m) (2c + f + x) + (2h + f + x)


Different members of the same family placed the following orders. Simplify the orders by combining like terms. (2h + f) + (c + f + s) + (h + m + f) = (x + c) + (2f + c + x) + (m + 2f + c) = (h + x + f) + (h + x + f) + (h + x + f) = (3h + m) + (2c + f + m) + (c + m + 2f) = (4c + f + m) + (3h + f + m) – h + c = (3h + 2f + x) + (c + f + m) – (h + m + f) = (5h + 3f + 2m) – (h + f + m) + (c + 2x) = (3h + f + x) + (h + f + m) – (h + f) + c =


Someone used the wrong letter in each of these orders. Can you solve the problem? h + 6w = $8.15 8x + 5y = 27.15 4b + 6l = 13.10 3m + 6p + 5f = 20.10 7h + 2f + 3c – 2g = 17.20 4(h + m) + 6u = 21.70 8c + 7(s + z) = 38.20


KEY h + 6w = $8.15 w = 1.05 (fries) 8x + 5y = 27.15 y = 2.15 (cheeseburger) 4b + 6l = 13.10 b = .95 (small soda) 3m + 6p + 5f = 20.10 p = 1.85 (hamburger) 7h + 2f + 3c – 2g = 17.20 g = 2.15 (cheeseburger) 4(h + m) + 6u = 21.70 u = 1.55 (large soda) 8c + 7(s + z) = 38.20 z = 2.05 (extra large soda)


Can you find the price of a hamburger and of an order of fries at each of these restaurants? Restaurant A: 3h + 2f = $11 h + 2f = $5 Restaurant B: 2h + 4f = $12 3h + f = $8 Restaurant C: What is the price of a cheeseburger? 2h + 3c + f = $20 3h + 2c + f = $19 h + 5c + 2f = $27


Something spilled on parts of these orders. What are the missing items?

Quantity Item Subtotal 3 fries $3.15 6 hamburgers 11.10 5 extra large 10.25 Total: $24.50

Quantity Item Subtotal 4 hamburger $7.40 2 cheeseburger 4.30 5 fries 5.25 2 small sodas 1.90 4 cheeseburgers 8.60 Total: $27.45


Menu Math Name_________________________ Evaluate each expression using the menu. Show your work below each problem.

8. h + f + m =

9. l + c + x + f = 3) 4h = 4) 2c + f = 5) 6f + 3c + 2x – 2m 6) 8x + 6m + 10c + 5l = 7) 4l + 4s + 4m = 8) 6c – 2c = 9) 6m + 3h + 2c – 4m = 10) 8f – 3h = 11) 3h – 4c = 12) 6x – 3s – 5c + 7h =

Hamburger ............ $1.85 Cheeseburger ........ $2.15 Fries ...................... $1.25 Sodas:

Small ................ $1.05 Medium ............ $1.35 Large ................ $1.65 Extra Large ....... $1.95


Distribute the following orders and combine like terms. 1) 4(h + m) = 2) 3(x + m + c) = 3) 7(2c) = 4) 5(3h + s) = 5) 8(2c + 4f + x) = 6) 4(3h + c + f + 5m) = 7) 13(2m + 2c – 4h) = 8) 7(3h – 5s + 12m) = 9) -4(3c + f) = 10) -6(4x + 2s – 3f) = 11) -5(7h – c – 9l) = 12) Evaluate problem number 4 by substituting the values from the menu. 13) Evaluate problem 7. 14) Which problems would have a negative value? 15) Which of these orders would not happen in reality?

Menu Math Name____________________________


Simplify these orders by combining like terms. When necessary, distribute first. 1) 6h + 3c + h = 2) 2c + 3x + 5m + 4x + m = 3) 6c + 2f + 4c + 3x + f + 2m + 6x = 4) 3(m + h) + 2m = 5) 7s + 4(h + 2l) = 6) 3m + 6(h + 2m) + 5h + x = 7) 4(h + f + m) + 3(c + f + l) = 8) 2(5x) + 3c + 4(x + c) = 9) 5h + 6m + 3l + f + (h + f) – 2m = 10) (c + x) + 3(h + x) + (2m + c) – h = 11) 7(h + 2m) + 2(x + h) + 3h + m – (h + x) = 12) (2c + s) + 4h + 4(h + f + m) – 3(h + s) = 13) Evaluate problem 8. 14) Evaluate problem 10. 7. Which two orders are identical?

Menu Math Name____________________________


Oops! Someone wrote the wrong letter on each order. Solve each equation to find out what the cook should make. Show your work neatly. 1) h + z = $2.90 2) 6d = $8.10 3) v + 2c = $5.35 4) 2r + 2f + m = $7.55 5) 3(c + s) + t + 2s = $13.35 6) 4p + 7(3c) = $50.15 7) 2(h + 2l) + b = $11.55 8) 4(h + 2x) – g = $21.35 9) 6(h + f) + 5n – 3h = $23.80 10) 4(c + w) = $14

Menu Math Name____________________________


If your students are a little younger, the following version of “Menu Math” should work well. They have been used in numerous 4th and 5th grade classrooms with great success. By simplifying the menu prices to reflect more kid-friendly decimals, even younger learners can grasp the concepts we expect of much older students. Bon appetit!


Menu Math

Hamburger .................................... $1.50 Cheeseburger ................................ $2.00 Fries .............................................. $1.20 Sodas: Small .......................................... $.75 Medium .................................... $1.25 Large ........................................ $1.50 Extra Large .............................. $2.25


Menu Math

Hamburger ..................................... Cheeseburger ................................. Fries ............................................... Sodas: Small ......................................... Medium ..................................... Large ......................................... Extra Large ..........


Name___________________ Week_____ Use your menu to evaluate each expression. • m + f =

• c + x =

• h + s + f =

• m + 2f =

• 2h + l =

• 3m + 2c + 4s =

• 4m + 2m + m =

• 3c + h + 3c =

• h + 3m + 3x + s =

• 2h + 2c + 2m + 3f + 2x =


Name___________________ Week_____

1) c + d = $3

2) x + y + s = $4.05

3) 10h + 6g = 19.50

4) 3h + r = $6.50

5) f + 2z = $5.25

6) 20m + 10a + 15c = $68

7) 2m + 2w = $4.60

8) 2(c + m) + p = $7.80

9) 2(m + 2h) + 2b =$10.60

10) 3(h + n) + c + 2x = $16.10

Oops! Someone wrote these orders incorrectly. Can you identify each mystery item? Use your menu to help you.


VanHiele research on geometry

he Dutch mathematicians Dina and Pierre VanHiele developed the seminal model on the acquisition of geometric understanding in the 1950’s. Though their findings have been validated and supported for decades, it has been slow to find its way into

the American education system. In elementary and even in middle school, geometry is often overly simplified when students are asked to memorize content without exploring and developing it. Other times it is passed over entirely, or merely relegated to the final chapter of the book – a no-man’s land where teachers rarely find time to venture. Thus for many students their first venture into the domain of geometry comes when they have to pass a high-level course in secondary school. This coupled with the fact that the part of the brain that is dedicated to geometry is not the same region that deals with numerical mathematics means that many students fail this required course. However, the solution to this problem is clear and straightforward. Students who are taught consistently through the VanHiele model are much more likely to develop the necessary skills to succeed in geometry. Level 0: Visualization

Children recognize shapes by appearance: square, circle, rectangle. A child may call a sphere or cylinder a circle at this point, not distinguishing between 2D and 3D shapes. For example, a coin is a circle to children at this level. Students may apply the term hexagon to an octagon. Similarly, if a shape does not fit their classification scheme, they may reject it. A rotated square may be called a diamond or rhombus. An hourglass or bowtie shape may not be called a hexagon because it is not regular. A student may not be able to identify the base of a trinagle that has a horizontal side at the top and a vertex pointing downward. These students see shapes as separate classifications and ignore their interrelationships. For example, they don’t see a rectangle as a subset of parallelograms. Many older students and even adults are at this level of geometric understanding. To move them beyond this stage, one good activity is the “This is/This isn’t” activity. Given a set of shapes, you could say, “This is a polygon,” or “This is not a polygon,” until students note the similarities and differences among them.

Level 1: Analysis

At this level, students will focus on the properties inherent in shapes. These students realize that a rotated square is still a square. The characteristics and properties of a shape take precedence over its appearance.

T


They will begin to define a square by its properties, though they may not be able to do this perfectly. They might say a square has four congruent sides and neglect the fact that it also has four congruent angles. To develop this stage, educators should expose students to activities that will illustrate the properties of shapes.

• Create any triangle and cut it out. Remove the vertices and set them upon a common point. How many degrees are there? (180º)

• Create any quadrilateral. Locate the midpoints of the side. Connect them to form a new quadrilateral. What is the name of this shape? (Parallelogram)

• Compare the diagonals of different quadrilaterals. What characteristics do they share?

Manipulative and computer-based activities are crucial. • https://www.geogebra.org/ • https://www.geogebra.org/m/FAhtKpR5 • https://www.geogebra.org/m/VkxdAZrG • https://www.geogebra.org/m/GFwZ5qdf#material/YT2AVyyp

Level 2: Abstraction

Students will begin to see how shapes relate to one another and can see that a square is therefore both a rhombus and a rectangle. They do this by seeing that properties of one shape may apply to another also. They will begin to reason about shapes and their properties, though this is often based on inductive reasoning (recognition and generalizations of patterns and similarities based on observations). To develop this level of ability, lead the students to make a discovery such as the fact that the vertices of a quadrilateral add up to 360º. Then have them test this repeatedly with various types of regular and irregular quadrilaterals. Although students at this stage of development show a high level of understanding, they fail to reason deductively or to understand the need for postulates, conjectures, and theorems. They follow hunches and intuition more than proof. Again, geometry software can be of great help in developing these generalizations.

Level 3: Deduction

This is the classic stage of high school geometry. They can reason deductively (based on absolute truths that can be proven). These students can follow or create a deductive proof given certain initial conditions. To help students in this stage, begin with the simplest of proofs. For example, if we accept that all triangles have an interior angle measurement of 180º, then we can prove that quadrilaterals must have an interior angle measurement of 360º since any quadrilateral can be divided into two triangles. Similarly, any pentagon can be


subdivided into three triangles for an interior angle sum of 540º. Continuing this way, it can be shown algebraically that the formula for the interior angle sum of any polygon is 180(n–2) where n represents the number of sides. It may be helpful to students if you compare this stage to a court trial. We cannot base guilt and innocence on hunches and simple observations of patterns: “The last three people who got caught speeding had red cars, so if you have a red car, you are guilty of speeding.” In a court proceeding, guilt must be proven, even if it is obvious. We depend upon evidence such as fingerprints or DNA that cannot be refuted. Though in most cases, inductive reasoning will get us through, there are times when we want to be absolutely sure. An astronaut going into space wants assurance that the rocket will get work there.

Level 4: Rigor

At this level, we can explore beyond plane geometry. For example, lines of latitude are perpendicular to the equator but don’t produce parallel lines. Instead they converge in both directions due to the curvature of the earth’s surface. We would also find more rigorous proofs at this level, such as proof by negation.

Sadly, though most students are at level 0, or at best, level 1, high school geometry is taught at levels 3 and 4. And unlike some subjects, students must proceed through these levels sequentially; they cannot skip steps and find success. It is best to imagine the five levels as rungs on a ladder. Students must have every rung in place to ensure they can reach the upper heights. Fortunately, to some degree the movement from one rung to the next is not dependent solely on age but is accelerated by experiences. That means that as we provide these opportunities to students in elementary and middle school, they are more likely to find success in high school geometry.


A B D c

E

F

G

H

I J

K L

M N O

P

Title:_________________________

Definition:__________________________________________________________________________ __________________________________________________________________________________

Level 0


A B D c

E

F

G

H

I J

K L

M N O

P

Title:_________________________

Definition:__________________________________________________________________________ __________________________________________________________________________________


Rectangles

Parallelograms

Rhombi

Level 1


Rectangles

Parallelograms

Rhombi


Make three points on the page and join them to create a triangle. Cut out the triangle and arrange the vertices around a common point. What is the degree sum of the three angles? Compare your results with your team.


Make three points on the page and join them to create a triangle. Cut out your triangle. Find the midpoints to two sides and join them. How does this median of the triangle compare the the third side?


Make four points on your paper and join them to form a quadrilateral. Find the midpoints of each side and connect them to form a new quadrilateral. What type of quadrilateral is the new shape? Compare your results with your team.


Graphic Organizers for Teaching Algebra Students who struggle mathematically often fail for one of two reasons: either they struggle with the arithmetic that they encounter or they struggle with what to do with that arithmetic. For example, in solving a two-digit by two-digit multiplication problem like the one shown here, the student must begin by finding the product of the six and the nine. Given enough fingers, toes, and time, most students can find that 6x9=54. However, the student must also know that the four goes beneath the nine, and the five goes above the four. Then the nine and four must be multiplied while the five is added and both numbers go next to the four. Then for some inexplicable reason, a mystery zero appears. The student may approach mathematics with the thought that, “Ours is not to reason why–just invert and multiply.” Clearly we want our students to understand why an algorithm works, but realistically few of them do. Just as importantly, we want our students to be successful with math, and few of our intervention students are. For this reason, if we can provide students with a structural template or a graphic organizer, they need only concern themselves with the arithmetic involved in a problem. Following are a few graphic organizers that will help students with common mathematical tasks. Lattice multiplication: For centuries, multiplication was solved differently than it is now. A lattice was be used to multiply 46 by 79 as shown in the left diagram. Each digit was multiplied and the product placed in the intersecting cells. The tens digit was placed above the diagonal and the ones was placed below. In the second diagram, the six and nine have been multiplied to get a product of 54. All cells have been filled out in the third diagram.

The

following diagram shows how the diagonals are added to produce the product. Notice that in two diagonals, a tens digit has been regrouped to the top of the next diagonal.

4 6

7

9

4 6

7

9 5 4

4 6

7

9 5

4

4 2

3 6

2 8


Although the lattice method does essentially the same steps as our more familiar process, I have observed that students who use this method tend to make fewer errors. I believe this is because they do not have to attend to the structure of the algorithm and simply focus on the arithmetic of the multiplication. This method of multiplication also transitions very nicely into multiplication of binomials in algebra. Rarely do we teach students to multiply binomials the way we multiply multi-digit numbers. Instead more and more algebra teachers are using what is called the generic rectangle that is essentially the lattice. If we want to multiply two binomials such as (x+3)(x-4) we can place them in an array. Again, we fill in each cell and add diagonals as we did in the lattice. Now however, we call it combining like terms.

4 6

7

9 5

4

4 2

3 6

2 8

4 3

6

3 1 1

3,634

x +3

x

-4

x2 +3x

-12 -4x

x2–x–12


Proportion Boxes: Solving proportions, percents, and the word problems involving them are easier with this graphic organizer. Let’s look at a sample word problem that lends itself to a solution involving proportions.

Notice that we are comparing two ideas in two ways. We are comparing the part of his day he spends sleeping to his whole day. We are also comparing a single day to his whole life. Thus we could label our graphic organizer as shown. This sets up the proportion in the proper arrangement. That is, we have 7/24=x/54. We would now decide the best method of solving this proportion. In this case, cross products would be effective. Notice that if the labels are switched, the numbers switch with them and a valid proportion still results. This template will work for any proportion problem, even if it is not a word problem. In addition, it works for all types of percent problems. Problems about percent can be solved using the same strategy. Simply change the labels as shown. To solve the problem, we simply put the numbers into their correct cells. What percent of 56 is 21? This gives us the proportion:

21/56 = x/100

Mr. Fulton sleeps 7 hours per day. How many of his 54 years has he spent sleeping?

hours years

part

whole

7 x

54 24

fraction percent

part

whole

21 x

100 56


Combined Mixture Problems: Combined mixture problems in algebra can baffle students. They tend to look like one of these two problems.

I have used the following approach with great success. First, I tell them these problems are “ca-ca”. They agree. Then I explain that they can be solved with an equation that contains a lot of “ca-ca”.

C1A1+C2A2=CTAT This means the concentration of the first solution times the amount of the first solution plus the concentration of the second solution times the amount of the second solution equals concentration of the total solution times the amount of the total solution. In the second example the c stands for cost instead of concentration. Before substituting values into the equation template, I introduce the graphic organizer shown here. The students must decide which numbers go in which cells. First I ask the students to think about the problem. Clearly, if we mix two solutions, the total concentration of the mixture will be stronger than one solution and weaker than the other. That is, the total concentration will be the middle percent. (This will also be true if a cost is involved.) Thus the 20% is the total, and the 10% and the 75% are the concentrations of the two addends. Then the students merely match the amounts with the respective percents and use an x for the missing number. The completed graphic is shown. Notice that the two amounts in the top cells must add up to the total amount. The template for the second problem is also shown. It is also a good idea to have the students estimate the answer ahead of time. For example, in the second problem, the total mixture is one dollar less than the $6 price of the walnuts but three dollars more than the $2 price of the peanuts. Thus we will need more of the walnuts than we will of the peanuts. Often students can use this information to correctly estimate the answer.

A chemist wants to mix 12 grams of a 10% salt solution with some 75% salt solution to get a 20% salt solution. How many grams of the 75% salt solution are needed? A company wants to create a 20 mixture of peanuts and walnuts that sells for $5 per pound. The peanuts sell for $2 per pound and the walnuts sell for $6 per pound. How many pounds of each must they use?

1

+2

T

C A

1

+2

T

C A

10% 12

75% x

20% x+12

1

+2

T

C A

$2 x

$6 20–x

$5 20


“X” Marks the Spot

Overview: The rules are so simple you can teach them without saying a word! Yet the math is rich and abundant. You can use these simple drills to reinforce basic addition, subtraction, multiplication, and division facts and develop number sense without boring your students. Fractions, decimals, and negative numbers can also be used. You can even factor polynomials using this simple method! 1. Tell the class that this game has only two simple rules...but you won’t tell them what

they are. They will have to figure out the rules by themselves. As soon as a student knows how to play, he or she can come up to the board and write down the answer.

2. Have them copy these five problems onto a piece of paper as you write them on the board.

Then begin writing in the answers by adding the numbers on the left and right to get the bottom number and multiplying them to get the top number.

3. If you work slowly, pausing as if to ponder before writing each answer, some of the

students will soon catch on. After a majority of the class has discovered the rules of the game, allow a student to explain them.

4. Then you can continue to play the game by varying the format. • Placing numbers in sections A and B will require students to

divide first, then add.

3 4 2 8 1 9 53

5 6 9

3 4 2 8 1 9 53

5 6 9 7

12

10

16

10

9

10

25

15

54

A

B

D

C

Materials: ý paper o activity master

5

10 10/5 = 2

5 + 2 = 7


• Placing numbers in sections A and D or C and D will

require students to subtract first, then multiply. • Placing numbers in sections B and D will require

students to study the various combinations of sums and products that satisfy the given answers.

5. Eventually, you may wish to increase the difficulty through examples like these.

6. You can also use the guess and check method to solve complex puzzles. Research has shown that the guess and check method is not only a valuable skill, it helps children transition to solving equations in algebra. Here is how to solve problems like the one on the right using this method. Pick a pair of numbers that add up to 100 such as 50 and 50. Write them in columns a and b. Then multiply them to find the product. In this case, it is 2500, which is too high. We mark our check with an "H" to signify that this is too high. This tells us that the number in column a is too high. Let's adjust our guess by trying 40 and 60. Remember that our guesses must add to 100. It is also very important to note that the smaller of the two numbers must go in column a.

1 x 20 = 20 1 + 20 = 21 2 x 10 = 20 2 + 10 = 12 4 x 5 = 20 4 + 5 = 20

20

9

12

8 12 – 8 = 4

4 x 8 = 32

15 -9 1.5 .25 1/2 7/8

6.25

5

-16

0

xy

x+y

6

0 16

0

2491

100

a b check

50 50 2500 H

40 60 2400 L

45 55 2475 L

48 52 2496 H

47 53 2491 J


The product of these two numbers is 2400, which is too low. This is marked with an "L". Our next guess for column a must be greater than 40 but less than 50. Let's try 45. This makes b = 55. Our new product, 2475 is too low also. Our fourth guess will be 48. Now b = 52, and our product is 2496. Although this is too high, it is very close. For our next guess, we try 47 for a, and 53 for b. This gives us the product we wanted.

× × × × Ø Ø Ø Ø

Journal Prompts: ? Explain to a student how you would find the solution to the problem on the

right. What could you tell about the value of a and c in the example on the below?

What can you tell about the value of b? Explain.

Homework: & Assign one of the accompanying activity masters. You can make a homework worksheet by placing numbers in a copy of the blank activity master. Alternately, the students can copy down problems as you write them on the board.

Taking a Closer Look: The difficulty of these drills can be varied by the numbers chosen

and their placement. Using decimals, fractions, or negative numbers can also increase the complexity.

Algebra students can practice factoring polynomials this way too. For the polynomial x2 + 7x + 10 = 0, students would construct the problem shown. I tell them to put the b term in the basement and the c term in the ceiling. The solutions are 2 and 5. The expression factors into the following binomials:

x2 + 7x + 10 = (x + 2)(x + 5) The solution to the equation then is x = –2 and x = –5.

48

19

Good Tip! C These drills are a great way to practice number concepts throughout the year. Worksheets can be created on the spot to be used as homework or warm-ups. If you are studying fraction multiplication, simply have the students copy a set of these problems as you write them on the board.

10

7


Assessment: þ These drills can be spot checked for accuracy or students can exchange papers to check

them. You may also use the answer keys for the accompanying activity masters.


Answer Key:

Set 1 top bottom Set 2 side top Set 3 side bottom Set 4 side side a 14 9 a 5 35 a 8 10 a 4 7 b 45 14 b 6 24 b 7 14 b 4 8 c 12 7 c 0 0 c 6 14 c 3 12 d 8 9 d 7 49 d 7 13 d 6 8 e 36 12 e 12 60 e 9 13 e 6 6 f 36 13 f 9 54 f 12 15 f 0 12 g 48 14 g 11 22 g 6 12 g 4 9 h 12 8 h 9 108 h 4 16 h 7 9 i 0 7 i 12 96 i 6 10 i 8 8 j 9 6 j 10 10 j 8 15 j 6 10 k 20 9 k 12 48 k 7 11 k 5 12 l 25 10 l 12 120 l 0 9 l 8 11 m 33 14 m 2 12 m 3 8 m 10 12 n 60 16 n 12 132 n 2 11 n 11 12 o 24 14 o 0 0 o 7 10 o 7 12 p 77 18 p 6 42 p 12 21 p 12 12 q 72 20 q 9 72 q 11 21 q 8 9 r 60 16 r 12 0 r 12 23 r 9 11 s 60 17 s 12 144 s 1 9 s 9 9 Set 5 top bottom Set 6 top bottom Set 7 side side Set 8 side side a 1 7.2 a -14 -5 a -36 -1 a 47 53 b 0.45 1.4 b 45 -14 b -1 36 b 23 63 c 1.2 4.3 c -12 1 c -6 6 c 50 60 d 0.008 0.18 d 8 -9 d 2 14 d 211 289 e 3.6 6.6 e -36 0 e -14 -2 f 0.36 1.3 f 36 -13 f -7 4 g 0.048 0.68 g -48 2 g -4 7 h 0.12 0.8 h -12 4 h -9 -2 i 0 0.7 i 0 -7 i -6 3 j 0.9 3.3 j 9 -6 j -6 -3 k 0.002 0.09 k -20 1 k -18 -1 l 0.025 0.55 l -25 0 l -9 -5 m 0.33 1.4 m 33 -14 m -15 -3 n 6 10.6 n -60 -4 n -15 3 o 0.24 2.12 o 24 -14 o -1 45 p 0.077 0.81 p 77 -18 p -8 -6 q 0.96 1.28 q -96 -4 q -6 8 r 0.6 10.06 r 60 -16 r -12 4 s 6 12.5 s -60 -7 s -2 24


“X” Marks the Spot Name_________________________

Multiply the two side numbers and put the product on the top. Add the two side numbers and put the sum on the bottom.


"X" Marks the Spot 1 Name________________________ Multiply the two side numbers and put the product on the top. Add the two side numbers and put the sum on the bottom as shown.

8 7

56

4 7 2 9 5 4 3

6 6 4 9 8 6 1 8

0 7 3 3 4 5 6 2

3 11 10 6 2 12 5 5

8 12 6 10 12 5 11 7

a b c

e f g d

i j k h

m n o l

q r s p

Activity Master


"X" Marks the Spot 2 Name________________________ The number on the bottom is the sum of the two numbers on the sides. Find the missing side number. Then multiply the two side numbers and write the product on the top.

8 7

56

15 7

12 4

10 9

9

5 17

15

6 2 13

20

8 1 11

16

4

8

6 23

11 8 8

8 17

0 12

24

12

14

7

21

12

10 22

13

7

a b c

e f g d

i j k h

m n o l

q r s p

Activity Master


"X" Marks the Spot 3 Name________________________ The top number is the product of the two numbers on the sides. Find the missing side number. Then add the two side numbers and write the sum on the bottom.

8 7

56

15 2

16

49

7

48

8

42

6

36

4

36

3

36

6

48

12

24

4

56

7

28

4

0

9

15

5

18

9

21

3

108

9

110

10

132

11

8

8

a b c

e f g d

i j k h

m n o l

q r s p

Activity Master


"X" Marks the Spot 4 Name________________________ The top number is the product of the two missing side numbers. The bottom number is the sum of the two missing side numbers. Find the missing side numbers.

8 7

56

15

28

11

32

12

36

15

36

12

0

12

36

13

64

16

60

16

60

17

120

22

132

23

84

19

72

17

99

20

81

18

48

14

63

16

88

19

144

24

a b c

e f g d

i j k h

m n o l

q r s p

Activity Master



.8 7

5.6

7.8 7 .2 .9 .5 4 .3

.6 6 .4 .9 .08 .6 .1 .08

0 .7 .3 3 .04 .05 .6 .2

.3 1.1 10 .6 2 .12 .5 .05

.08 1.2 .06 10 12 .5 .11 .7

a b c

e f g d

i j k h

m n o l

q r s p

Activity Master



-8 7

-56

-1 -7 2 -9 -5 4 -3

6 -6 -4 -9 8 -6 -1 -8

0 -7 -3 -3 -4 5 6 -2

-3 -11 -10 6 -2 -12 5 -5

8 -12 -6 -10 -12 5 -11 -7

a b c

e f g d

i j k h

m n o l

q r s p

Activity Master


"X" Marks the Spot 7 Name________________________ The top number is the product of the two missing side numbers. The bottom number is the sum of the two missing side numbers. Find the missing side numbers.

1 36

36

37

36

-37

-36

35

-36

0

28

16

28

-16

-28

-3

-28

3

18

-11

-18

-3

18

-9

18

-19

45

-14

45

-18

-45

-12

-45

44

48

-14

-48

2

-48

-8

-48

22

a b c

e f g d

i j k h

m n o l

q r s p

Activity Master


"X" Marks the Spot 8 Name________________________ Use a guess and check table to find the missing side numbers. Always put your lower number in column a.

2491

100

a b check 1449

86

a b check

3000

110

a b check 60979

500

a b check

a

d c

b

Activity Master


Using “X” Marks the Spot to Solve Combined Work Problems.

One of the more challenging types of problems for algebra students is the combined work problems. These often look like these two examples:

Andy can paint a room in three hours. Zoe can paint the same room in four hours. How long would it take them to paint the room working together? One pipe can fill a tub in three hours. A second pipe can fill the same tub in four hours. How long will it take to fill the tub if both pipes are used?

Typically students either add the two numbers or average them, failing to realize that the job must get done more quickly than either given time. Thus the answer must be less than either given number. The most common way to solve the problem is by using the following formula:

The variable, x, represents the total time for the combined work. The first fraction means that Andy paints one third of a room per hour, and the second fraction shows that Zoe paints one fourth of a room per hour. The two fractions add up to one room being painted. While this procedure makes sense to the teacher, its origin and development is beyond the ability of most students. With enough practice, students may become proficient with this, but it remains questionable if the equation holds meaning for them. Let’s look at the solution to this problem using the equation.

Now examine the solution to the following problem from “X” Marks the Spot:

Notice that the solution to this problem is also 12/7! Certainly this is a much simpler way to achieve the correct answer, but will it always work? In fact, it does. Let’s demonstrate this by generalizing the problem:

Andy paints a room in a hours, and Zoe paints the same room in b hours. How long will it take them to paint the room working together?

4 3


This is also the solution to the simpler problem when we use a and b as our side numbers: This method will also work when only one person’s time is given along with the combined work time:

One pipe can fill a tub in 3 hours. With a second pipe running, the tub can be filled in only two hours. How long would the second pipe take if it was used alone?

The traditional algorithm would look like this:

Using the “X” Marks the Spot strategy, we have: We also know that the answer is two. Thus the top number divided by the bottom number must equal two:

b a a+b

ab

x 3 x+3

3x


Multiplying binomials using the box In this activity, students use a 2x2 box to multiply binomials. It is much like filling in a multiplication table. Each cell of the box contains the product of the factors above and to the left of it. Multiply (2x+1)(3x–4)

Two worksheets are provided. In the first, the a term is 1, and in the second it is greater than 1.

2x +1

6x2 – 5x – 4

3x

–4

6x 2

–8x

3x

–4


Multiplying Name______________________________

Binomials Date___________________ Class_______ Use the area model to multiply these binomial pairs.

x +4

(x + 4)(x + 2) = ____________

x

+2

A x +7

(x + 7)(x + 9) = ____________

x

+9

B x +3

(x + 3)(x + 8) = ____________

x

+8

C

x – 4

(x – 4)(x + 2) = ____________

x

+2

D x +5

(x + 5)(x – 9) = ____________

x

–9

E x – 6

(x – 6)(x + 1) = ____________

x

+1

F

x – 8

(x – 8)(x – 4) = ____________

x

–4

G x – 2

(x – 2)(x – 7) = ____________

x

–7

H x – 9

(x – 9)(x – 7) = ____________

x

–7

I


Multiplying Name______________________________

Binomials Date___________________ Class_______ Use the area model to multiply these binomial pairs.

2x +1

(2x + 1)(x + 1) = _____________

x

+1

A x +5

(x + 5)(4x + 3) = _____________

4x

+3

B 3x +1

(3x + 1)(2x + 1) = _____________

2x

+1

C

3x – 2

(3x – 2)(3x + 4) = ____________

3x

+4

D 4x – 5

(4x – 5)(2x – 9) = ____________

2x

– 9

E 6x – 5

(6x – 5)(5x + 1) = ____________

5x

+1

F

7x – 8

(7x – 8)(8x + 7) = ____________

8x

+7

G 6x – 1

(6x – 1)(6x – 5) = ____________

6x

– 5

H 2x – 11

(2x – 11)(4x – 3) = ___________

4x

– 3

I


Factoring polynomials where the a term is 1 using the X Now students will use what they learned about using “X Marks the Spot” to factor trinomials in which the a term is 1. The b term goes in the basement. We put the product of the a and c terms in the attic and ceiling.

To factor, simply solve as we did with “X Marks the Spot”. Factor x2 + 3x – 10 a = 1, b = 3, c = –10 ac = –10

x2 +3x – 10 = (x – 2)(x + 5)

The attic or ceiling. Put the product of the a and c terms here

The b term goes in the basement.

–10

3


Factoring Name______________________________

Polynomials Date___________________ Class_______

Use the X to factor each polynomial. Write the answer in factored form as in the example.

12

7 4 3 x2 + 7x + 12 = (x + 3)(x + 4)

x2 + 11x + 28 = ____________________

28

11

A

x2 + 14x + 33 = ____________________

33

14

B

x2 – 8x + 16 = ____________________

16

-8

C

x2 – 8x – 20 = ____________________

-20

-8

D

x2 + 31x – 32 = ____________________

-32

31

E

x2 – 13x + 40 = ____________________

40

-13

F


x2 + 3x – 28 = ____________________

G

x2 + 16x + 48 = ____________________

H

x2 – 8x – 48 = ____________________

I

x2 – 19x + 48 = ____________________

J

x2 – 3x – 54 = ____________________

K

x2 + 25x – 54 = ____________________

L

x2 – 19x – 42 = ____________________

M


Factoring polynomials when a >1 using the box or the X and box In the first of the two worksheets, students factor trinomials when a > 1 using the box. This requires some trial and error. In the second worksheet, students combine the power of the X and box to solve these challenging polynomials with no guesswork. First we recall that in the box method, the a term goes in the upper left cell, and the c term goes in the lower right. The be term is the sum of the two remaining cells. Remember also that when using the X, the product of the a and c terms goes in the attic-ceiling and the b term goes in the basement. Let’s factor 12x2 – x – 6. We fill in the X and box with what we know. a = 12, b = –1, c = –6 ac = –72

a

c

ac

b

12

–6

–72

–1


Solving the X gives us 8 and –9. These are placed into the remaining cells of the box. It doesn’t matter in which order they are placed. Now we simply factor the columns and rows. Notice that the bottom row has a common factor of 3. However, since the lead term (–9x) is negative, the factor is also negative. The factorization of 12x2 – x – 6 is (3x + 2)(4x – 3).

–72

–1

8 –9 12x 2

–9x

8x

–6

3x +2

4x

–3

12x 2

–9x

8x

–6


Challenging Name______________________________ Polynomials Date___________________ Class_______

Use a box to factor each polynomial. Write the answer in factored form as in the example.

2x +5

2x2 + 9x + 10 = (2x + 5)(x + 2)

x

+2

x2 + 8x + 15 = ________________

A

2x 2

4x

5x

+10

x 2

+15

x2 – 5x – 14 = ________________

B

x2 + 4x – 32 = ________________

C

x 2

– 14

x 2

– 32

2x2 + 3x + 1 = ________________

D

3x2 + 7x + 2 = ________________

x

+2

E

2x 2

+1

3x 2

+2


9x2 – 6x + 1 = ________________

H

6x2 + 7x + 2 = ________________

I

4x2 – 9 = ________________

J

4x2 – x – 3 = ________________

K

8x2 + 2x – 1 = ________________

L

8x2 + 6x – 5 = ________________

M

+1

2x2 + 7x + 3 = ________________

x

F 3x

3x2 + 5x – 2 = ________________

+2

G

2x 2

+3


Using the Name______________________________ “X” and Box Date___________________ Class_______ Use the X and box to factor each polynomial as in the example. Remember to write in the correct attic, ceiling, and basement numbers. Write your answer in factored form.

18

7 – 2 9

6x2 + 7x – 3 = (2x + 3)(3x – 1) 2x +3

3x

– 1

6x 2

– 2x

9x

– 3

A. _______________=(______)(______)

B. _______________=(______)(______)


C. _______________=(______)(______)

D. _______________=(______)(______)

E. _______________=(______)(______)

F. _______________=(______)(______)


Using the Name______________________________ “X” and Box Date___________________ Class_______ Use the X and box to factor each polynomial as in the example. Remember to write in the correct attic, ceiling, and basement numbers. Write your answer in factored form.

18

7 – 2 9

2x +3

3x

– 1

6x 2

– 2x

9x

– 3

– 20

8

A. 4x2 + 8x – 5 = (______)(______)

4x 2

– 5

– 12

1

B. 4x2 + x – 3 = (______)(______)

4x 2

– 3

6x2 + 7x – 3 = (2x + 3)(3x – 1)


C. 4x2 – 13x + 3 = (______)(______)

4x 2

+3

D. 8x2 – 2x – 15 = (______)(______)

E. 6x2 + 35x – 6 = (______)(______)

F. 20x2 – 8x – 1 = (______)(______)


Answer Keys

Multiplying Binomials 1 A) x2 + 6x + 8 B) x2 + 16x + 63 C) x2 + 11x + 24 D) x2 – 2x – 8 E) x2 – 4x – 45 F) x2 – 5x – 6 G) x2 – 12x + 32 H) x2 – 9x + 14 I) x2 – 16x + 63 Factoring Polynomials 1 A) (x + 4)(x + 7) B) (x + 3)(x + 11) C) (x – 4)(x – 4) D) (x – 10)(x + 2) E) (x – 1)(x + 32) F) (x – 8)(x – 5) G) (x – 4)(x + 7) H) (x + 4)(x + 12) I) (x – 12)(x + 4) J) (x – 16)(x – 3) K) (x – 9)(x + 6) L) (x – 2)(x + 27) M) (x – 21)(x + 2)

Challenging Polynomials 1 A) (x + 3)(x + 5) B) (x – 7)(x + 2) C) (x – 4)(x + 8) D) (2x + 1)(x + 1) E) (3x + 1)(x + 2) F) (2x + 1)(x + 3) G) (3x – 1)(x + 2) H) (3x – 1)2

I) (3x + 2)(2x + 1) J) (2x + 3)(2x – 3) K) (4x + 3)(x – 1) L) (4x – 1)(2x + 1) M) (4x + 5)(2x – 1) Using the X and Box 1 A) (2x + 5)(2x – 1) B) (4x – 3)(x + 1) C) (4x – 1)(x – 3) D) (4x + 5)(2x – 3) E) (x + 6)(6x – 1) F) (10x + 1)(2x – 1)


Notes


Seminar Evaluation Please rate the seminar by circling the appropriate numbers below: 1 (low) 7 (high)

Content of the seminar 1 2 3 4 5 6 7

Contribution of the instructor 1 2 3 4 5 6 7

Content of the resource manual 1 2 3 4 5 6 7

The seminar as a whole 1 2 3 4 5 6 7

Something I plan to use: ...................................................................................................

Something I’d suggest: ..................................................................................................

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