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Lesson Plans and Web-Based Tutorials for College Factoring Workshops

by

JOHN N BRUNNER

December 2010

A Project Submittedin Partial Fulfillment of

the Requirements for Degree of

MASTER OF SCIENCE

The Graduate Mathematics ProgramCurriculum Content Option

Department of Mathematics and StatisticsTexas A&M University-Corpus Christi

APPROVED: ___________________________________ Date: ____________ Dr. George Tintera, Chair

___________________________________ Dr. Elaine Young, Member

___________________________________ Dr. Sherrye Garrett, Member

Style: APA

Abstract

The process of factoring polynomials can be a difficult skill to master,

especially for beginning algebra students at the college level. One of the many

basic skills that a developmental mathematics student is exposed to and must

comprehend is the process of factoring. This process involves finding an

equivalent expression that is a product (Bittinger, Ellenhogen, & Johnson, 2006).

Without a strong background in basic mathematical skills, students can easily

develop frustration and disillusionment, hindering their potential to be successful

college graduates.

A local community college provides support for students enrolled in

developmental mathematics courses, including workshops, supplemental

instruction, and one-on-one tutoring. Factoring workshops are taught twice a

semester, providing support for students enrolled in remedial level mathematics

classes. At the present time, no written lesson plans or videos are provided by

the mathematics department for facilitators of the workshops.

The purpose of this project was to develop eleven detailed lesson plans

on topics presented in the factoring workshops as well as extensive web-based

tutorials on factoring polynomials by use of video recordings. These lesson plans

will be integrated into future workshops for use by presenters or student tutors.

The video tutorials will allow students unlimited access for viewing on the

college’s web site if additional instruction is needed.

ii

Dedication

For my inspiration and understanding I give full thanks to Dr. Nadina

Hutchings. Before you entered my life, I used to wonder what I might accomplish.

You will always be in my heart, my thoughts, and I miss you so very dearly.

Acknowledgements

I would like to thank my committee chair Dr. George Tintera for all his

patience, assistance and caring, and Dr. Elaine Young for her professional

guidance and the wonderful aspects of mathematics you have taught me so that I

can become a better teacher in the future. I would also like to thank Dr. Sherrye

Garrett; through your course of 2008 I received inspiration to read again and

ponder how a child learns.

I would finally like to thank my family and friends, especially Faydale Curtice

for the many hours of assistance you have provided me throughout this project. I

am always truly grateful for your patience and caring. To my dear friends that I

met while attending graduate school, Nimisha Bhakta, Sonia Carrera and

Conchita Marshall, I am truly blessed with such wonderful people.

Table of Contents

iii

Abstract.................................................................................................................. ii

Dedication............................................................................................................. iii

Acknowledgements............................................................................................... iii

Introduction............................................................................................................5

Literature Review..................................................................................................7

Results................................................................................................................11

Summary.............................................................................................................15

References..........................................................................................................16

Appendix A: Complete Lesson Plans..................................................................20

Appendix B: Student Packet..............................................................................133

Appendix C: Student Packet Answer Key.........................................................155

Appendix D: Hyperlinks for Videos....................................................................161

Appendix E: Additional Problems and Solutions Available for Students............162

iv

Introduction

Some students enter the higher education environment without the fundamental

mathematical proficiency and problem solving skills that are needed to succeed in a

college level mathematics course (Hammerman, 2003). Without a strong background in

the basic mathematical skills, students do not succeed in higher level courses such as

calculus or linear algebra, nor will they be eligible for many high-technology and high-

paying fields after graduation (Paglin & Rufolo, 1990). A local community college

provides one-on-one tutoring, access to use of computers for online homework

assignments, extended operating hours six days a week, and implementation of

factoring workshops, which are held twice a semester.

Attending the workshops is a diverse array of the school population, ranging from

elementary algebra to college algebra level students. The workshops are presented on

Saturdays during the first two weeks of the fall and spring semesters, to coincide with

the start of classes. These workshops usually last about five hours, with an hour lunch

and intermittent breaks for students and instructors. Workshops are held again during

the last two weeks of the semester to assist students with their first exposure to

factoring techniques while taking elementary algebra classes.

One area of concern is the lack of structured lesson plans for use by the

instructors or fellow tutors. Previous workshops have been somewhat disorganized with

a “shoot from the hip” attitude. This conveys to the students a diminished level of

professional development, as well as may confuse the remedial level students even

further in their quest to understand the factoring process used in mathematics.

5

The overall purpose of this project was to develop eleven lesson plans covering

the topics that are discussed in the factoring workshops. In addition, sensing a need to

“fill the gap” for those that are not able to attend the workshops in person due to work

schedules, concurrent classes or family matters, the college’s audio visual facilities and

the author of this project produced a series of videos mirroring the content of the lesson

plans. Similar problems presented at the workshops were worked out, step-by-step, and

were shown on video by use of document camera capturing. These tutorials will be

available for the students to view on the college’s website for additional instructional

support.

This project also developed additional problems, answers, and detailed solutions

to five randomly selected problems from each of the ten topics. These problems will be

posted on the college’s website, in print form only, for viewing by students, if they so

choose.

This project was guided by two main principles:

1) Factoring in mathematics is extremely difficult and requires extended learning

opportunities which can be provided by the workshops.

2) On line tutorials can provide additional support to students unable to attend

the workshops.

6

Literature Review

At the local community college, the decentralized developmental mathematics

program consists of the following three courses: Arithmetic and Geometry, Elementary

Algebra, and Intermediate Algebra. These developmental mathematics courses

normally serve multiple purposes. The primary goal is to strengthen student deficiencies

in basic mathematical skills which are needed to succeed in other college math classes,

as well as courses that require some form of math, such as chemistry, economics or

engineering (National Association of Developmental Education, 2002 ). At many

community colleges, developmental courses also serve a second purpose of

strengthening students’ overall general learning skills before they enroll in regular

college courses.

Remedial or developmental education classes, such as the ones offered at the

local community college, are defined as coursework below college-level. Developmental

education is a topic of much heated debate in higher education (National Council of

Postsecondary Research, 2008). Colleges and states devote substantial resources to

try and resolve the issue of remediation. In addition, students bear the cost of enrolling

in the remedial level courses, with the possibility of lost wages and prolonged school

time in order to graduate. In the U.S., almost a quarter of postsecondary students

required remedial course work in mathematics in the fall of 2000, with almost 98% of

public two-year colleges and 80% of four-year institutions offering at least one remedial

mathematics course (National Council of Educational Statistics, 2003).

A national initiative on community college success studied over 46,000 students

enrolled in 27 institutions reported over 70% of students were referred to developmental

7

mathematics courses versus only 34% for developmental English (Biswas, 2007). The

local community college in this project, from fall 2004 to fall 2008, reported 61% to 89%

of first time students tested into a remedial level mathematics course, compared with

only 30% to 59% for developmental English and only 27% to 46% for developmental

reading courses (Institutional Research and Effectiveness, 2008).

In order for a person to be successful in any mathematics course, one must have

“appropriate math knowledge” (Nolting, 2002, p. 42). To determine this mathematics

knowledge, community colleges or universities require students to take placement

exams in order to determine their current level of mathematics knowledge. These test

scores are then used to place students into appropriate courses. Students that were

successful in their recommended developmental courses tend to continue their

enrollment in college for a greater number of terms (Stutz & McCarroll, 1993). The

problem is that students disregard these placement suggestions and enroll in classes

that are above the recommended level, or take the placement exam multiple times to

avoid taking lower level mathematics courses (Nolting, 2002). This practice can lead to

frustration and feelings of being overwhelmed when they enroll in the wrong class.

Hollis (2009) recommends academic advisors become an integral part of the process to

help developmental level students overcome their learning obstacles while attending

college and to guide them in determining what level of classes should be taken.

Nolting (2002) asserts, “Learning math is different from learning many other

subjects because it follows a sequential learning pattern” (p. 22). Sequential learning

patterns are based on knowledge or skills acquired in certain mathematics classes will

be used again in the future to build a stronger mathematical foundation. These

8

sequential learning patterns can be strengthened if students enroll in appropriate

classes using placement exams as well as scheduling mathematics courses

sequentially and continuously in order to maintain continuity in the upper level

educational learning process (Nolting, 2002).

The community college in this project incorporates remedial level assistance in

the three following content areas: reading, writing, and mathematics. The student

tutoring center is one of several key components in the overall process to assist

students who are taking remedial level courses. It provides one-on-one tutoring, access

to an adequate number of computers in order for students to work on assignments,

extended operating hours six days a week, supplementary handouts, course content

mastery (other than mathematics) and most importantly, factoring workshops.

Of the three developmental mathematics courses offered, the topics of factoring

are covered in the last two courses. Factoring, in its simplest terms, poses the question,

“What was multiplied together to obtain the given result?” (Bittinger, Ellenbogen, &

Johnson, 2006, p. 304). In other words, to factor in mathematics is to find an equivalent

expression that is a product (Hall, 2010). Terms are a single expression, called a

monomial; a two-term expression is called a binomial; a three-term mathematical

expression is considered a trinomial, and anything larger than three terms is labeled a

polynomial (Bittinger, Ellenbogen, & Johnson, 2006).

However, the process of factoring in elementary and intermediate algebra can be

a difficult skill to comprehend and master for students in remedial level classes

(Kotsopoulos, 2007). This skill is understood quickly by some students, but others take

a longer time to grasp the concept. One reason may be the diversity of learning styles

9

and multiple intelligences that are present in the classroom setting (Gawlik, 2009).

To assist students taking remedial level classes that are actively covering the

topics of factoring, further allocation of resources should be provided by means of

additional practice, extended study time, additional personnel, or a different teaching

method (Hoda, 2006).

Computer use and all its encompassing technologies has become a normal part

of a person’s daily routines. “Internet technologies make learning environments

available without restrictions in time or place.” (Pahl, 2002, p. 1). Online courses and

distance modules are growing rapidly in science, mathematics, engineering, and

technology disciplines (Hauk, 2006; National Science Foundation, 1998). Technology

has provided new ways of bringing instructional delivery to the student. The use of

electronic tutorials will allow the student to review missed work and materials (Handal &

Herrington, 2003). Each year more web-based tutorials are being used to deliver course

work through distance learning, as well as to supplement traditional classroom settings.

Evidence comparing the use of web-based tutorials show that academic achievement

using web-based tutorials is equal to traditional classroom instruction (Singleton &

Fernandez, 2005).

The use of technology, such as the web-based tutorials developed for this

project, allowed students to learn the topics of factoring at their own pace, replay a

tutorial as many times as needed to understand the concept, meet their own learning

styles, and be exposed to a different teaching method. The additional resources

available by use of web tutorials assist in engaging the student’s multiple intelligences

(Roubides, 2004).

10

Results

The lesson plans developed for this project were initiated in July 2010 and

concluded in August 2010. They include:

1) Tests for divisibility by the natural numbers 2 through 10

2) Greatest common factors

3) Factoring by grouping method

4) Sum of perfect squares

5) Difference of perfect squares

6) Sum of perfect cubes

7) Difference of perfect cubes

8) Monic trinomials: x2+bx+c

9) Non-monic trinomials: a x2+bx+c a≠1 ,0

10) Techniques of factoring

11) Solving for specified variable

The lesson plans were developed in this order to maintain organization and

create a level of sequential learning patterns (Nolting, 2002). Rules of divisibility will be

used in later topics such as monic and non-monic trinomials, thus is presented first in

the sequence.

A lesson plan template was chosen from www.docstoc.com, mainly due to

simplistic layout and author’s membership. Some alteration of wording was done to fit

the college level setting. Each lesson plan consisted of the following subcomponents:

1) Title

2) Student learning objective

11

3) Standards addressed

4) General goals

5) Required materials

6) Anticipatory set (Lead in)

7) Step-by-step procedures

8) Plan for independent practice

9) Closure (reflect anticipatory set)

10) Examples

For the step-by-step procedures section, problems used in the lesson plans were

chosen from the author’s five personally owned textbooks. The numbers of problems

chosen for each lesson plan ranged from 4 to 6 with detailed explanations being

presented for the benefit of the instructor.

Incorporating from five personally owned textbooks, this author also randomly

selected problems of varying degree of difficulty, developed listings of problems, and

their respective answers for the purpose of additional support for students at the

community college. After consultation, the author and an academic advisor initially

agreed to provide ten additional problems from each of the major topics, excluding

divisibility rules and techniques of factoring, with detailed step-by-step solutions. Upon

implementing this strategy, it became quickly evident that the length of project would

become too great. Therefore, the decision was made to curtail the number of detailed

problems to only five. It is important to note that the problems worked out from each

topic consist of detailed procedures, with each step scrutinized closely to maintain a

12

logical flow for the sake of the remedial level students. This author feels strongly that

deliberate concise steps shown on paper or on video will assist in the learning process.

The number of problems selected for each section ranged from as low as 30 to

as high as 78. All problems or equations were created using Microsoft Office 2010®,

implementing the mathematics equation editor feature. This created a highly

professional appearance throughout the entire project.

The lesson plans will be available for future use by this author or by fellow tutors

as the need arises, and will be kept at the tutoring center as soon as approval of this

project is completed. The assortment of problems, answers and detailed solutions to the

five selected problems from each section will be converted to PDF format for placement

on the community college’s website as soon as approval of this project is completed.

Video recordings using the community college’s audio-visual facilities began in

late October 2010, with sessions lasting approximately three to four hours, with two

sessions each week, ending in the middle of November 2010. The recordings mirrored

the lesson plans and types of problems presented at the workshops to maintain

continuity if viewed by remedial level students at home or school. Many technical

obstacles concerning sound quality and lighting were faced during the first two weeks of

filming. The initial strategy in filming was to use a combination of a document camera

capturing coupled with the use of a whiteboard. This did not materialize due to poor

lighting in the room, which created a distracting “glare” over the dry erase marks on the

whiteboard. Upon further contemplation and brainstorming, this author used the

Microsoft Office 2010® mathematics equation editor feature to create the largest font

numbers and equations possible and used the document camera feature to show step-

13

by-step how to factor correctly and to solve certain equations. The camera did not

capture “live time”, but was producing still images about every four to five seconds. This

technique of using large font numbers and equations worked very well after some

adjustments were made to speech rate and tone of delivery.

The video recordings were of a standing position only, using an external audio

microphone attached to lapel of a shirt. This approach was an excellent way to convey

body language and seemed to produce a more dynamic presentation. The author

became very comfortable in front of video camera by the third week of recordings. The

video tutorials will be available on the community college’s website after approval of this

project is completed.

A student packet was also developed, encompassing the following:

1) Formulas for factoring, presented in two formats

2) Tips on factoring polynomials

3) List of perfect square and perfect cube numbers

4) List of first thousand prime numbers

5) Rules of divisibility

6) Graphic organizer on factoring polynomials

7) Rules of divisibility worksheet

8) Five sample problems from each topic

9) Answer key for student packet problems

This packet will be distributed at beginning of workshops and will be available

online for further use by students as desired.

14

Summary

The process of factoring polynomials can be a difficult skill to master, especially

for beginning algebra students at the college level. A local community college provides

support for students enrolled in various developmental mathematics courses, including

workshops, supplemental instruction, one-on-one tutoring and content mastery (other

than mathematics). Factoring workshops are taught twice a semester, providing support

for students enrolled in remedial level mathematics classes. At the present time, no

written lesson plans are provided by the mathematics department for instructors of the

workshops.

The purpose of this project was to develop eleven detailed lesson plans on the

topics presented in the factoring workshops and extensive web-based tutorials on

factoring polynomials by use of video recordings. These lesson plans will be integrated

at future workshops for use by presenters or fellow tutors. The video tutorials will allow

students unlimited access for viewing on the college’s web site, if additional instruction

is needed.

Presentation of factoring workshops combined with integration of lesson plans

that were developed for this project will assist in the students’ comprehension and

understanding of a very important topic that is used in mathematics on a regular basis.

Development of this project has furthered the professionalism of this author

greatly since inception earlier this year. In addition, remedial level students at the

community college will benefit with additional resources available, either in print form or

video.

15

References

Bittinger, M. L., Ellenbogen D. J., & Johnson, B. L. (2006). Elementary and Intermediate

Algebra. New York: Addison Wesley.

Biswas (2007). Accelerating remedial math education: How institutional innovation and

state policy interact. Retrieved November 25, 2010 from

http://www.achievingthedream.org/_pdfs/_publicpolicy/RemedialMath.pdf

Gawlik, C. (2009). Making effective video tutorials: An investigation of online written and

video help tutorials in mathematics for preservice elementary school teachers.

Unpublished doctoral dissertation, Kansas State University. Retrieved March 15,

2010 from http://hdl.handle.net/2097/1608

Hall, E. (2010). Elementary Algebra. Corpus Christi, TX: Author.

Hammerman, N. (2003). Strategies for developmental math at the college level. Math

and Computer Education, 37(1), 79-95.

Handal, B., & Herrington, A. (2003). Re-examining categories of computer-based

learning in mathematics education. Contemporary Issues, 3(3), 275-287.

Hauk, S. (2006). Web-based homework in college algebra: A comparison of web-based

and paper and pencil homework on student performance in college algebra.

Retrieved April 14, 2010, from

http://hopper.unco.edu/faculty/personal/hauk/segalla/WBWquan_060307.pdf

Hoda, N. E. (2006). Evaluation of effects of a curriculum-based math intervention

package with elementary school-age students in a summer academic clinic.

Doctoral dissertation, Mississippi State University, 2006. Retrieved March 10,

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http://hdl.handle.net/2097/1608

2010, from http://sun.library.msstate.edu/ETD-db/theses/available/etd-07312006-

143107/unrestricted/Dissertation.pdf

Hollis, L. (2009). Academic advising in the wonderland of college for developmental

students. College Student Journal, 43(1), 31-35. Retrieved from ERIC database.

Institutional Research and Effectiveness. (2008). Developmental course enrollments.

Retrieved March 21, 2010, from http://www.delmar.edu/IRE/

Kotsopoulos, D. (2007). Unravelling student challenges with quadratics: A cognitive

approach. Australian Mathematics Teacher, 63(2), 19-24.

National Association for Developmental Education. Best practices in developmental

mathematics. Mathematics Special Professional Interest Network. Retrieved

November 25, 2010 from

http://www.math.csi.cuny.edu/Faculty/CourseDevelopment/MTH015/bestpractice

s.pdf

National Council of Postsecondary Research. The impact of postsecondary remediation

using a regression discontinuity approach: Addressing endogenous sorting and

noncompliance. A NCPR working paper. Juan Carlos Calcagno & Bridget Terry

Long (2008). Retrieved November 25, 2010 from

http://www.postsecondaryresearch.org/i/a/document/8162_CalcagnoLongRevise

d.pdf

National Center for Education Statistics (NCES)(2003). Remedial education at degree

granting postsecondary institutions in fall 2000. Washington, D.C. U.S.

Department of Education. Retrieved April 10, 2010 from

http://nces.ed.gov/pubs2004/2004010.pdf

17

http://sun.library.msstate.edu/ETD-db/theses/available/etd-07312006-143107/unrestricted/Dissertation.pdf

http://sun.library.msstate.edu/ETD-db/theses/available/etd-07312006-143107/unrestricted/Dissertation.pdf

National Science Foundation. (1998). Information technology: Its impact on

undergraduate science, mathematics, engineering, and technology (NSF Report

Number NSF98-82). Retrieved March 15, 2010 from

http://www.nsf.gov/pubs/1998/nsf9882/nsf9882.txt

Nolting, P. (2002) Winning at math. Your guide to learning mathematics through

successful study skills. Bradenton, FL: Academic Success Press, Inc.

Paglin, M., & Rufolo, A. M. (1990). Heterogeneous human capital, occupational choice,

and male-female earnings differences. Journal of Labor Economics, 8(1), 123–

44.

Pahl, C. (2002). An evaluation of scaffolding for virtual interactive tutorials. In M. Driscoll

& T. Reeves (Eds.), Proceedings of World Conference on E-Learning in

Corporate, Government, Healthcare, and Higher Education 2002 (pp. 740-746).

Chesapeake, VA: AACE.

Retrieved November 25, 2010 from http://www.editlib.org/p/15295

Roubides, P. (2004). The use of technology in enhancing students’ learning. In J. Nall &

R. Robson (Eds.), Proceedings of World Conference on E-Learning in Corporate,

Government, Healthcare, and Higher Education 2004 (pp. 903-909).

Chesapeake, VA: AACE.


Singleton, A. M. K. & Fernandez, E. (2005). Exploring the effect of student learning

styles on learning using a web-based tutorial. In P. Kommers & G. Richards

(Eds.), Proceedings of World Conference on Educational Multimedia,

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Hypermedia and Telecommunications 2005 (pp. 4074-4079). Chesapeake, VA:

AACE.


Sturtz, A. J., & McCarroll, J. A. (1993). Placement testing and student success: The first

intervening variable. Paper presented at the Annual Forum of the Association for

Institutional Research. Chicago, IL. Retrieved November 28, 2010 from http://0-

web.ebscohost.com.portal.tamucc.edu/ehost/resultsadvanced?

vid=4&hid=106&sid=5cd0062a-ed07-4a40-b54f-c37eef8dd65c

%40sessionmgr115&bquery=(Placement+testing+AND+student+success+

%3a+the+first+intervening+variable)&bdata=JmRiPWVyaWMmdHlwZT0xJnNpd

GU9ZWhvc3QtbGl2ZQ%3d%3d

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Appendix A: Complete Lesson Plans

Lesson Plan 1: Divisibility Rules

Instructor Notes

20

Lesson Plan Title: Divisibility Rules

Student Learning Objectives: Students learn the rules of divisibility for the numbers 2,

3, 4, 5, 6, 7, 8, 9 and 10. Students use these rules to check large numbers for their

divisibility, and to determine factors for monic and non-monic trinomials.

Standards Addressed: Texas College and Career Readiness Standards

I. Numeric Reasoning

A. Number representation

1. Compare real numbers.

B. Number operations

1. Perform computations with real numbers.

C. Number sense and number concepts

1. Use estimation to check for errors and reasonableness of solutions.

II. Algebraic Reasoning

A. Expressions and equations

1. Explain and differentiate between expressions and equations using words

such as “solve,” “evaluate,” and “simplify.”

General Goal(s): Instructor greets students at the workshop, discusses topics to be covered, and

advises that even though a factoring workshop is being held, the first topic to be

discussed is to teach students about certain divisibility rules. The divisibility rules will be

used in later sections of the workshop, such as determining greatest common factor,

factoring by grouping method, and eventually solving for a certain variable. Instructor

advises students divisibility really means one factor is determined. A numbern is

21

divisible by5, for example, really meansn=5∗someother factor ; we are trying to find

someother factorwhen using the divisibility rules.

Required Materials: Erasers (brought by students)

Paper (brought by students)

Pencils (brought by students)

Worksheets (provided by facilitator of workshop)

Anticipatory Set (Lead-In):

As students arrive at workshop, hand out student packet.

Step-By-Step Procedures:

Dividing by 2

1. All even numbers are divisible by 2. That is, all numbers ending in 0,2,4,6 or 8.

Dividing by 3

1. Add up the digits in the number.

2. If the sum is divisible by 3, so is the number.

3. For example: Look at the number 1230. Just add up the digits: (1+2+3+0) = 6.

The instructor asks the students, “Is 6 divisible by 3?” “Yes it is.” So that means

the number 1230 is also divisible by 3.

4. This process may have to be repeated for larger numbers.

Dividing by 4

1. Just look at the last two digits in the number. If the last two digits are divisible by

4, then that number is also divisible by 4.

22

2. For example: 2048. Look at the last two digits only. Instructor asks the students,

“Is 48 divisible by 4?” “Yes, it is.” So that means the number 2048 is also divisible

by 4.

Dividing by 5

1. Numbers ending in a 5 or a 0 are always divisible by 5. Mention to the students

the aspect of money when seeing numbers ending in a 5 or 0.

Dividing by 6

1. Check for divisibility by 2 and divisibility by 3. Recall, if the number is even, then

it is divisible by 2. In addition, if the sum of the digits is divisible by 3, then the

number is also divisible by 3. If a number is both divisible by 2 and 3, then the

number is also divisible by 6. Stress to students that a number must be divisible

both by 2 and 3 for that number to be divisible by 6. If a number is not divisible by

one or the other, then the number is not divisible by 6.

2. For example: 162. This number is even, so it is divisible by 2. In addition, the

sum of the digits is equal to 9. We know 9 is always divisible by 3. Therefore, 162

is also divisible by 3. Since 162 is divisible by 2 and also by 3, it is also divisible

by 6.

Dividing by 7

1. Take the last digit in a number.

2. Double and subtract the last digit, which is the ones digit, from the rest of the

given number.

3. You may have to repeat the process for larger numbers.

23

4. If the result after subtracting is divisible by 7, then the given number is also

divisible by 7.

5. Example: 539 (Double the 9 to get 18. Subtract 18 from 53 to get 35, which is

divisible by 7. We can now say that 539 is divisible by 7.

Dividing by 8

1. If the last 3 digits are divisible by 8, so is the entire number.

2. Example: 1088. Just look at the last three digits. Instructor covers up the 1 with a

finger. Just look at 088. The last 3 digits are divisible by 8, therefore, so is 1088.

Dividing by 9

1. The rule for divisibility by 9 is similar to rule for divisibility by 3.

2. Add up the digits in the number.

3. If the sum is divisible by 9, so is the number.

4. Example: 1287. Add up (1+2+8+7) = 18. The instructor asks the students “Is 18

divisible by 9?” “Yes, it is.” Therefore, 1287 is also divisible by 9.

Dividing by 10

1. If the number ends in a 0, it is divisible by 10.

2. Example: 67890. The ones digit is 0, therefore, 67890 is also divisible by 10.

Example:

Check 555 for divisibility by 2, 3, 4, 5, 6, 7, 8, 9, or 10.

1) Divisible by 2: No, since 555 is not an even number.

2) Divisible by 3: Yes. (5+5+5) = 15. “Is 15 divisible by three?” “Yes, it is.” Therefore,

555 is also divisible by 3.

3) Divisible by 4: No, since 55 is not divisible by 4.

24

4) Divisible by 5: Yes, since the number ends in a 5.

5) Divisible by 6: No, since 555 is not an even number. Both rules for divisibility by 2

and divisibility by 3 must apply.

6) Divisible by 7: No.

7) Divisible by 9: No, since the sum of the digits (5+5+5 = 15). “Is 15 divisible by 9?”

“No, it is not.” So therefore, 555 is not divisible by 9.

8) Divisible by 10: No, since 555 does not end in a 0.

Plan for Independent Practice:

Worksheet contained in student packet provided by workshop.

Closure (Reflect Anticipatory Set): Instructor advises students in workshop that

divisibility comes into play a great deal of the time when dealing with trinomials,

determining the LCD, and working with the greatest common factor type problems.

25

Name: Date:

Divisibility Rules

Number Divisible By My Hypothesis The Actual Rule

2

3

4

5

6

7

8

9

10

Russell, D. (n.d.)

26

Name___________________

Divisibility Practice

Determine if each number is divisible by 2, 3, 4, 5, 6, 7, 8, 9, or 10.

Make a list under each number.

1.) 210 2.) 72 3.) 145

4.) 333 5.) 1400 6.) 62

7.) 1008 8.) 2310 9.) 144

10.) 231

Russell, D. (n.d.)

27

Answer Key:

1.) 2, 3, 5, 7

2.) 2, 3, 4, 6, 8, 9

3.) 5

4.) 3, 9

5.) 2, 4, 5, 7, 8, 10

6.) 2

7.) 2, 3, 4, 6, 7, 8, 9

8.) 2, 3, 5, 6, 7, 10

9.) 2, 3, 4, 6, 8, 9

10.) 3, 7

References

Bittinger, E. (2006). Elementary and Intermediate Algebra. Concepts and Applications.

Boston: Pearson.

Hall, E. (2010). College Algebra. Corpus Christi, TX: Author.


Hall, E. (2010). Intermediate Algebra. Corpus Christi, TX: Author.

Russell, D. (n.d.).www.about.com.Retrieved July 15, 2010 from

http://math.about.com/library/bldivide.htm

Sullivan, M. (2002). College Algebra. Upper Saddle River: Prentice Hall.

Swokowski, C. (1993). Fundamentals of College Algebra. Boston: PWS Publishing

Company.

28

Texas Higher Education Coordinating Board. (2008). Texas College Readiness

Standard. Retrieved March 21, 2010 from http://www.thecb.state.tx.us/

End of lesson plan

29

Lesson Plan 2: Greatest Common Factors

Instructor Notes

30

Lesson Plan Title: Greatest Common Factors

Student Learning Objective: Application of factoring and solving polynomials,

specifically factoring out the greatest common factor(s) from two or more terms.




1. Explain and differentiate between expressions and equations using

words such as “solve,” “evaluate,” and “simplify.”

B. Manipulating expressions

1. Recognize and use algebraic (field) properties, concepts, procedures,

and algorithms to combine, transform, and evaluate expressions (e.g.,

polynomials, radicals, rational expressions).

C. Solving equations, inequalities, and systems of equations


and algorithms to solve equations, inequalities, and systems of

linear equations.

2. Explain the difference between the solution set of an equation and the

solution set of an inequality.

D. Representations

1. Interpret multiple representations of equations and relationships.

2. Translate among multiple representations of equations and

relationships.

31

General Goal(s): Identify and factor completely polynomials paying close attention to

greatest common factor(s) of two or more terms. Students will be able to demonstrate

that factoring polynomials having common factors is the opposite operation of

multiplying a polynomial by a monomial. Students in workshop will be able to find the

greatest common factor(s) of two or more terms.

Specific Objectives: Do operations with polynomials, specifically factoring out the

common factor, number, variable or complete expression that might be contained within

a series of mathematical terms.




Worksheets (provided by facilitator of workshop)Anticipatory Set (Lead-In): Instructor would lead into discussion of one of the first

skills of factoring is dealing with the greatest common factor, and its importance in

higher level mathematics classes. The one basic skill that is seen time and time again in

all types of math problems is the process of factoring out the greatest common factor(s)

from two or more terms. This invariably makes further computations easier and the

equation or problem much more manageable.


1) The instructor ask the audience the meaning in mathematics of the following

phrase: “What is a product and what is a factor?”

32

2) Instructor discusses with students the following: To sum in math means to

add. The expressions to be added are called terms. On the other hand, to find a product

means to multiply in math. The numbers or variables being multiplied are called factors.

Example: 6ab (x+1)

This expression is considered to be in factored form. The6abis considered a

factor as well as the expression( x+1 ).

Given the problem: 10 x2(2a−5 x)

How many factors do we have? What are the factors? Instructor discusses with

students the following expression is in factored form. The first factor that is seen is 10 x2

and the second factor is(2a−5x ).

At this time, instructor will begin discussion of determining the greatest common

factor (GCF). Pose the question to the audience: “Are we able to decompose large

numbers or terms into smaller, more manageable, more easily seen common factors?”

“Is there a number, or variable that is a common factor to all terms in an algebraic

expression?”

Example: 2ax+2ay

1) Stress importance of being able to rewrite the following expression into the

following by decomposing the two terms into factors. Decomposing in math is to “break

down” a number or expression into smaller components. For these lesson plans

contained in this project, decomposing means to “factor out.”

2a∗x+2a∗y

33

Notice how the terms can be rewritten with the greatest common factor seen

more easily. Both terms have in common2a.

2) If instructor needs to use a different colored pen, pencil or dry erase marker to

make more distinction in order to show students the common factor, then that is

allowable. Notice the greatest common factor (GCF) is2a.

3) The greatest common factor is written down first, then to right of common

factor, we open a parenthesis to record what remains in each term followed by ending

parenthesis.

4) 2a(x+ y )

5) At this time, the instructor will rewrite answer without red mark.

6) 2a(x+ y )

Example:2 x2+2 x−8

1) 2∗x2+2∗x−2∗4 Decompose, or factor each term.

2) 2∗x2+2∗x−2∗4 GCF is 2.

3) 2(x2+x−4 ) GCF written first, followed by remaining

factors for each term.

4) 2(x2+x−4 )

The instructor advises students at this time, the three term expression contained

within the parenthesis is not factorable. Thex2+ x−4expression is considered a prime

polynomial.

Example: 4 a2+12a

34

1) 4 a∗a+4a∗3 Decompose, or factor each term.

2) 4 a∗a+4a∗3 GCF is 4 a.

3) 4 a(a+3) GCF written first, followed by remaining

factor for each term.

4) 4 a(a+3)

Example: x3 ( x+3 )−7(x+3)

1) Instructor will discuss with students that the GCF may also be an entire

expression enclosed within parenthesis, so( x+3 ) is a common factor.

2) x3 ( x+3 )−7(x+3) Use colored pen to make

distinction for common factor.

3) (x+3)(x3−7) GCF written first, followed by remaining

factors for each term.

4) (x+3)(x3−7)


Worksheet contained in packet provided by workshop.

Closure (Reflect Anticipatory Set): Instructor advises students in workshop that

greatest common factor type problems come into play a great deal of the time when

dealing with trinomials, determining the LCD, working with rational expressions, and

solving expressions for a certain variable. This skill is used throughout mathematics on

a regular basis. It is considered a fundamental skill or foundational skill that is common

to all level of mathematics.

35

References


Boston: Pearson.

Factoring polynomials having common factors. (n.d.). www.lessonplanet.com. Retrieved

October 31, 2010 from http://blue.utb.edu/jtelese/example%20lessons.htm






Company.



End of lesson plan

36

Lesson Plan 3: Factor by Grouping Method

Instructor Notes

37

Lesson Plan Title: Factor by grouping method

Student Learning Objective: To be able to rewrite trinomial as four term polynomial

and factor correctly. Factor completely if given a four term polynomial.














1. Recognize and use algebraic (field) properties, concepts, procedures, and

algorithms to combine, transform, and evaluate expressions (e.g., polynomials,

radicals, rational expressions).



algorithms to solve equations, inequalities, and systems of linear equations.

38

2. Explain the difference between the solution set of an equation and the solution

set of an inequality.

D. Representations


2. Translate among multiple representations of equations and relationships.

General Goal(s): Understand how to factor by using the grouping method, this usually

deals with four terms, but can be extended to trinomials also.





Anticipatory Set (Lead-In): Discuss with students that one technique of factoring

involves dealing with trinomials, which are three terms and also with four terms, which

are considered a polynomial. Grouping is one of the techniques that can be used

extensively to simplify polynomials. Instructor shows audience how a four term

polynomial such as the following, x2+2x+4 x+8 can be rewritten as(x+2)( x+4). Discuss

with students factoring by grouping deals with trinomials and four term polynomials.


x2+bx+c

1) The student must ensure the trinomial or polynomial is in correct order with highest

power of exponent first, and then the powers reduce as you read the trinomial or

polynomial from left to right.

39

2) Find two numbers whose product isc, BUT sum or difference isb.

3) The instructor and the students are trying to find two numbers that when multiplied

together will producec, but if added or subtracted will equal the middle coefficient. If that

is true, then rewrite the trinomial into a four term polynomial, decomposing or “breaking

down” only the middle term.

4) Factor the first pair of terms by enclosing in a set of parenthesis and do same for

second pair of terms.

5) Factor out the common monomial factor from each grouped set of parenthesis.

6) Write as a pair of binomials.

Example: x ²+9 x+14

Ensure terms are in correct order.

a) Find factors ofc, with sum or difference that equalsb.

Paired factors of 14 Sum Difference

14∗1 14+1=15 14−1=13

7∗2 7+2=9 7−2=5

Instructor asks audience what two numbers when multiplied together will produce

a product of 14, but if added or subtracted might produce the middle coefficient of 9.

Those two unique numbers are 7 and 2. At this point, we will rewrite the trinomial as

a four term polynomial, decomposing or “breaking down” the middle term of9 xinto

7 x∧2 x .

b) Group first two terms, group last two terms.

c) Factor out GCF from the two binomials.

40

d) Instructor will have two terms with common factor present contained within the

parenthesis.

e) Common factor will be written down first, followed by two remaining factors.

Instructor advises students to write problem down again on paper before

proceeding.

At this time, it is important for instructor to advise students that organization is

important to successful factoring. One way to assist in organization for students is to

recopy problem from above and give justification for action.

Problem Justification

1) x2+9 x+14 Rewrite problem.

2) x2+2 x+7 x+14 Rewrite as a four term polynomial. 3)

x∗x+2∗x+7∗x+7∗2 Decompose, or factor each term.

4) x∗x+2∗x+7∗x+7∗2 GCF’s are colored for more distinction.

5) x (x+2 )+7 (x+2) Factor out GCF’s.

At this point, instructor advises students the factor of( x+2 )is considered a

common factor and should only be written down once, even though it is seen twice

on paper. Proceed as follows:

6) (x+2)( x+7) Common factor written first, followed by

remaining two monomial factors.

7) (x+2)( x+7) Final answer.

Instructor can use different colored pens, chalk, or dry erase markers to make

the GCF’s more distinct for students. After much practice, the students will see the

GCF’s mentally, and use of different colored pens might not be an issue.

41

Example: x ²−6 x−55

Ensure terms are in correct order.

a) Find factors ofc, with sum or difference that equalsb.


55∗1 55+1=56 55−1=54

11∗5 11+5=16 11−5=6

Instructor asks audience what two numbers when multiplied together will produce

a product of 55, but if added or subtracted might produce the middle coefficient of 6.

Instructor advises students to not be overly concerned with sign in front of the middle

term. The signs will fall into place at a later time. Those two unique numbers are 11

and 5. At this point, we will rewrite the trinomial as a four term polynomial,

decomposing or “breaking down” the middle term of6 xinto−11 x∧5 x .

b) Group first two terms, group last two terms.

c) Factor out GCF from the two binomials.

d) Instructor will have two terms with common factor present contained within the

parenthesis.

e) Common factor will be written down first, followed by two remaining factors.

Instructor advises students to write problem down again on paper before

proceeding.


1) x2−6 x−55 Rewrite problem.

42

2) x2−11 x+5x−55 Rewrite as a four term polynomial. Instructor

advises students to think of the middle terms as−11 x+5x=−6 x.

3) x∗x−11∗x+5∗x−5∗11 Decompose, or factor each term.

4) x∗x−11∗x+5∗x−5∗11 GCF’s are colored for more distinction.

5) x (x−11)+5(x−11) Factor out GCF’s.

At this point, instructor advises students the factor of( x−11 )is considered a

common factor and should only be written down once, even though it is seen twice

on paper. Proceed as follows:

6) (x−11)(x+5) Common factor written first, followed by

remaining two monomial factors.

7) (x−11)(x+5) Final answer.

Instructor can use different colored pens, chalk, or dry erase markers to make

the GCF’s more distinct for students.

“How does a student check his or her work?” Do FOIL! Instructor mentions to

students to undo process of factoring, perform FOIL.

FOIL stands for:

First

Outer

Inner

Last

43

Factoring by grouping method extended for four terms

This technique of factoring by grouping can be extended to cubic equations or

cubic problems, as well.

1) Group first two terms, group last two terms.

2) Factor out GCF from two binomials.

3) Instructor will have two terms with common factor present contained within the

parenthesis.

4) Common factor will be written down first, followed by two remaining monomial

factors.

5) Write as a pair of binomials.

Example: x3+3x2+4 x+12


1) x3+3 x2+4 x+12 Rewrite problem.

2) (x¿¿3+3 x2)+(4 x+12)¿ Group terms.

3) (x2∗x+3∗x2)+(4∗x+4∗3) Decompose, or factor each term.

4) (x2∗x+3∗x2)+(4∗x+4∗3) GCF’s colored for more distinction.

5) x2 ( x+3 )+4(x+3) Factor out GCF’s.

At this time, instructor advises students that the( x+3 )expression is a common

factor and should be written down first, followed by remaining monomial factors.

6) (x+3)(x2+4) Common factor written first, followed by

remaining monomial factors.

7) (x+3)(x2+4) Final answer.

44

Stress to students that the two binomial expressions are completely factored at

this time, and no further manipulations are required. Instructor advises students to be

aware of sum of squares and difference of squares in final answers of problems. A sum

of squares, such asx2+4is not factorable. This topic of sum of squares will be discussed

in more detail in later lesson plans.

Example: 12 p3−21 p2+28 p−49


1) 12 p3−21 p2+28 p−49 Recopy problem.

2) (12 p3−21 p2)+(28 p−49) Group terms.

3) (3 p2∗4 p−3 p2∗7 )+(7∗4 p−7∗7) Decompose, or factor each term.

4) (3 p2∗4 p−3 p2∗7 )+(7∗4 p−7∗7) GCF’s colored for more distinction.

5) 3 p2 ( 4 p−7 )+7(4 p−7) Factor out GCF’s.

At this time, instructor advises students that the(4 p−7 )expression is a common

factor and should be written down first, followed by remaining monomial factors.

6) (4 p−7)(3 p2+7) Common factor written down first,

followed by monomial factors.

7) (4 p−7 )(3 p2+7) Final answer.

Stress to students that the two binomial expressions are completely factored at

this time, and no further manipulations are required.


Worksheet in student packet provided at workshop.

Worksheet for instructor use, if so desired:

45

Factor completely using grouping method:

1. x3+3x2+2 x+6

2. 6 z3+3 z2+2 z+1

3. 3 x3+2 x2+3x+2

4. 2 x3+6 x2+x+3

5. 8 x3−12 x2+6 x−9

6. 10 x3−25 x2+4 x−10

7. 12 p3−16 p2+3 p−4

8. 18 x3−21x2+30 x−35

9. 5 x3−5 x2−x+1

10. x3+8 x2−3x−24

46

Answer key to factoring by grouping method:

1. (x2+2¿ (x+3)

2. (3 z2+1 ¿(2 z+1)

3. (2 x2+1¿(x+3)

4. (x2+1¿ (3x+2)

5. (4 x2+3¿(2x−3)

6. (5 x2+2¿(2 x−5)

7. (4 p2+1¿ (3 p−4 )

8. (3 x2+5¿ (6 x−7)

9. (5 x2−1¿ (x−1)

10. (x2−3¿(x+8)

References


Boston: Pearson.






Company.



End of Lesson Plan

47

Lesson Plan 4: Sum of Perfect Squares

Instructor Notes

48

Lesson Plan Title: Sum of a Perfect Square

Student Learning Objective: Students will be able to understand and see the sum of a

perfect square type problem, determine its factorability or not.





2. Define and give examples of complex numbers.


1. Perform computations with real and complex numbers.














49



D. Representations



VII. Functions

A. Recognition and representation of functions.

1. Recognize whether a relation is a function.

2. Recognize and distinguish between different types of functions.

B. Analysis of functions

1. Understand and analyze features of a function.

2. Algebraically construct and analyze new functions.

C. Model real world situations with functions

1. Apply known function models.

2. Develop a function to model a situation.

VIII. Problem Solving and Reasoning

A. Mathematical problem solving

1. Analyze given information.

2. Formulate a plan or strategy.

3. Determine a solution.

4. Justify the solution.

5. Evaluate the problem-solving process.

B. Logical reasoning

50

1. Develop and evaluate convincing arguments.

2. Use various types of reasoning.

C. Real world problem solving

1. Formulate a solution to a real world situation based on the solution to a

mathematical problem.

2. Use a function to model a real world situation.

3. Evaluate the problem-solving process.

IX. Communication and Representation

A. Language, terms, and symbols of mathematics

1. Use mathematical symbols, terminology, and notation to represent given and

unknown information in a problem.

2. Use mathematical language to represent and communicate the mathematical

concepts in a problem.

3. Use mathematics as a language for reasoning, problem solving, making

connections, and generalizing.

B. Interpretation of mathematical work

1. Model and interpret mathematical ideas and concepts using multiple

representations.

2. Summarize and interpret mathematical information provided orally, visually, or

in written form within the given context.

C. Presentation and representation of mathematical work

1. Communicate mathematical ideas, reasoning, and their implications using

symbols, diagrams, graphs, and words.

51

2. Create and use representations to organize, record, and communicate

mathematical ideas.

3. Explain, display, or justify mathematical ideas and arguments using precise

mathematical language in written or oral communications.

X. Connections

A. Connections among the strands of mathematics

1. Connect and use multiple strands of mathematics in situations and problems.

2. Connect mathematics to the study of other disciplines.

B. Connections of mathematics to nature, and real world situations.

1. Use multiple representations to demonstrate links between mathematical and

real world situations.

2. Understand and use appropriate mathematical models in the natural, physical,

and social sciences.

3. Know and understand the use of mathematics in a variety of careers and

professions.

General Goal(s): For students to learn how to determine a sum of perfect squares by

inspection only.

Required Materials:

Erasers (brought by students)



Colored Pencils (brought by students)


52


Start with asking several questions directed toward audience:

1. What is a perfect square?

2. Would a number multiplied times itself produce a perfect square?

3. How can we visually recognize perfect squares?

4. What is the difference between a perfect square number and a square root?

5. What conclusion can you make about perfect squares?

For elementary level and intermediate level algebra students, instructor advises a

sum of squares is not factorable over the real numbers. This is considered a prime

polynomial, and cannot be broken down into two smaller binomials.

College algebra students only:

Solve forx:

1) x2+36=0

2) Instructor advises college algebra students only that the equation is already

set equal to zero and further computations can be performed.

3) At this time, instructor advises students to perform opposite operation of

adding 36, that is, subtract 36 from both sides of equation.

x2+36−36=0−36

4) Remind students to maintain equality by performing same operation to both

sides of equals sign. If you subtract 36 from the left side of equals sign, you must also

subtract 36 from right side of equals sign.

5) Proceed as follows:

x2=−36

53

6) At this time, discuss how to undo a squaring operation to audience. Pose the

question: “How does a person undo a squaring operation?” The student needs to solve

forx, notx2, therefore, the instructor advises students to take the square root of both

sides of equation. Proceed:

√ x2=±√−36

x=±√−36

7) At this time, remind students of the following rule:

Principal Square Root

√−r for r>0

Definition

√−r=i√r

Swokowski (1993)

Definition: A complex number is of the form a+bi where a∧bare considered real

numbers, and also √−1=i (Hall, E. 2010).

8) Instructor reminds college algebra students a few more steps are required in

order to finish problem, and that is to replace√−1withi.

x=± i√36

9) Instructor advises students to be familiar with a perfect square number such

as36.

x=±6 i

Closure (Reflect Anticipatory Set): Questions from audience taken at this time.

Questions that can be posed to audience: “Can students from memory recite squares of

natural numbers up to 20?" That is, 202=400.

54

Instructor discusses with audience that a sum of squares for elementary and

intermediate level algebra is not factorable. It is considered a prime polynomial. For

example, x2+4 y2 x2is a sum of squares that can be factored intox2 (1+4 y2 ) ,but the one of

the factors is still a sum of squares. For college algebra students, a sum of squares can

be solved, which leads to complex numbers.

References


Boston: Pearson.

Bravo, C. Factoring Polynomials. (2010). Hot Chalk’s Lessonplanspage.com. Retrieved

July 30, 2010 from

http://www.lessonplanspage.com/MathFactoringPolynomialsSquaresAndCubes9

12.htm






Company.



End of Lesson Plan

55

Lesson Plan 5: Difference of Perfect Squares

Instructor Notes

56

Lesson Plan Title: Difference of Perfect Squares

Student Learning Objective: An algebraic expression is a perfect square when the

numerical coefficient (the number in front of the variables) is a perfect square and the

exponents in the variables are even numbers.




















57



D. Representations



General Goal(s): Students will:

Analyze and represent patterns with symbolic rules.

Represent and compare quantities with integers.

Required Materials:





Anticipatory Set (Lead-In): Instructor briefly discusses with students the importance of

knowing perfect square numbers by inspection only when doing factoring of any type.

Can the student see squared numbers in any problem? Instructor at this time advises

students to view table of squared numbers supplied in student packet.


Note from previous lesson that the sum of two squares DOES NOT factor. The

difference of perfect squares is factorable and is seen numerous times in mathematics.

Advise students that memorization of all formulas is critical for successful factoring

skills.

58

Questions that can be posed to audience to determine if a difference of squares

is present:

1) How many terms do we have?

2) Is there a minus sign in between the two terms?

3) What is the power of the variable or the number?

4) Is that power to an even number?

Formula for Difference of Perfect Squares

F2−L2=(F−L ) (F+L )

F=first term L=last term

Example 1: Factor 9 x2−4

First note that there is no GCF to factor out of this binomial. The variable xis

raised to an even power, also 9∧4are considered perfect square numbers. A minus

sign is present between the two terms. This fits the form of a difference of perfect

squares. So we will be using the difference of perfect squares formula.

Instructor advises students to decompose, or factor each term separately.

3 x∗3 x−2∗2

This can be rewritten in a more compact form as follows:

¿

The instructor advises students that first term located within first set of

parenthesis is3 x. The last term located within second set of parenthesis is2.

F=3 x L=2

At this time, the instructor will insert first and last term into formula. The formula

for difference of squares is as follows:

59

(F+ L)(F−L)

(3 x+2)(3 x−2)

Note that if we multiply this out, that is, by doing the FOIL process, we would get

the original binomial. To undo factoring, simply perform FOIL!

Example 2: Factor (a+7¿2−16

First note that there is no GCF to factor out of this binomial. The term(a+7 )is

raised to an even power, also16is considered a perfect square number. A minus sign is

present between the two terms. This fits the form of a difference of perfect squares. So

we will be using the difference of perfect squares formula.

Instructor advises students to decompose, or factor each term separately.

(a+7 )∗(a+7 )−4∗4

This can be rewritten in a more compact form as follows:

(a+7 )2−( 4 )2

The instructor advises students that first term located within first set of

parenthesis is(a+7). The last term located within second set of parenthesis is4.

F=(a+7)L=4

At this time, the instructor will insert first and last term into formula. The

difference of perfect squares formula is as follows:

(F+ L)(F−L)

( (a+7 )−4 ) ( (a+7 )+4 )

One more step is needed to finish problem. Proceed as follows by removing

innermost parenthesis within both expressions.

60

(a+7−4)(a+7+4 )

(a+3 )(a+11)

Note that if we multiply this out, that is, by doing the FOIL process, we would get

the original binomial. To undo factoring, simply perform FOIL!

Closure (Reflect Anticipatory Set): Instructor mentions to students for difference of

squares, two terms will be present. A minus sign will be positioned between the two

terms, and the exponents of the variable or the number(s) will be raised to an even

power.

Plan for Independent Practice: Worksheet contained in packet provided at workshop.

Sample problems for instructor to use, if so desired:

Part A

Write only final product:

1) ( x+9 ) ( x−9 )=¿

2) ( y+z ) ( y−z )=¿

3) (6 x−1 ) (6 x+1 )=¿

4) (3 y+7 ) (3 y−7 )=¿

5) (x3−2 ) (x3+2 )=¿

6) ( xy−10 ) ( xy+10 )=¿

7) (x y2+z3 ) (x y2−z3 )=¿

8) (xn+ ym ) (xn− ym )=¿

9)( z4−15 ) ( z4+15 )=¿

10) ( xyz−4 ) ( xyz+4 )=¿

61

Part B

Factor Completely:

1) x2−100

2) y2− 1100

3) 1−4 z2

4) 25m2−9n2

5) x6−36

6) 25 r2−49 s2

7) 81 x4− y 4

8) 16 x4− ( y−2 z )2

9) a8−256

10) x2−( y−z )2

Part A answers:

1) x2−9

2) y2−z2

3) 36 x2−1

4) 9 y2−49

5) x6−4

6) x2 y2−100

7) x2 y 4−z6

8) x2n− y2m

9) z8−225

62

10) x2 y2 z2−16

Part B answers:

1) (x−10)(x+10)

2) ( y−1

10)( y+ 1

10)

3) (1−2 z )(1+2 z )

4) (5m−3n)(5m+3n)

5) (x3−6)(x3+6)

6) (5 r−7 s)(5 r+7 s)

7) (9 x2− y2)(9 x2+ y2)

(3 x− y )(3 x+ y )(9 x2+ y2)

8) (4 x2− y+2 z )(4 x2+ y−2 z )

9) (a4−16)(a4+16)

(a2−4 )(a2+4)(a4+16)

(a−2)(a+2)(a2+4)(a4+16)

10) (x− y+z )(x+ y−z)

References


Boston: Pearson.





63


Company.



End of Lesson Plan

64

Lesson Plan 6: Sum of Cubes

Instructor Notes

65

Lesson Plan Title: Sum of Cubes

Student Learning Objectives: Application of factoring and solving polynomials,

specifically two term expressions, which are called binomials. Recognize by inspection

a sum of cubes.








and algorithms to combine, transform, and evaluate expressions (e.g.,

polynomials, radicals, rational expressions).



and algorithms to solve equations, inequalities, and systems of linear

equations.



D. Representations


66


relationships.

General Goal(s): Identify and factor completely binomials that are sum of cubes.

Specific Objectives: Perform operations with polynomials, specifically two term

expressions, which are called binomials.





Anticipatory Set (Lead-In): Instructor again would lead into discussion of perfect cube

numbers with the students. Instructor refers students to table of perfect cubed numbers

in chart format enclosed in student packet.


A. Common methods of factoring polynomials are based on the number of its terms:

1. For any number of terms, factor out the greatest common factor(s) first.

2. For four or more terms, group like terms and factor out common factors.

3. For two terms:

Factor the sum of two cubes into the product of a binomial sum and

quadratic trinomial whose second term is negative.

Factor the difference of two cubes into the product of a difference binomial

and quadratic trinomial with all positive terms.

Advise students that memorization of all formulas is critical for successful

factoring skills.

67

Questions that can be posed to audience to determine if a sum of cubes is

present:

1) How many terms do we have?

2) Is there a plus sign in between the two terms?

3) What is the power of the variable or the number?

4) Is that power to a multiple of three?

Sum of Cubes

The formula for the sum of two cubes is: A3+B3=(A+B)(A2−AB+B2)

A=first termB=last term

Some instructors prefer this format:F3+L3=(F+L)(F2−FL+L2)


At this time, instructor discusses with students the importance of recognizing

perfect cube numbers, variables, or complete expressions. Some examples are

27 , x9 , ( x+7 )6.

Example 1: x3+27

For this type problem, two terms are present, which implies a binomial, a plus

sign is inserted between the two terms and the exponent is cubed. In addition, the

number 27 is considered a perfect cube. This leads to the assertion that a sum of cubes

is present. Decompose, or factor each term separately. Proceed as follows:

x∗x∗x+3∗3∗3

The instructor and students can put this into more compact form.

68

( x )3+ (3 )3

At this time, the instructor advises the students what the first term and last term

are.

first term=x

last term=3

Following the format of the sum of cubes formula, instructor advises to insert

terms. For the benefit of the students, write down formula on board:

(F+ L)(F2−FL+L2)

( x+3 )(( x )2−3 x+(3 )2)

One more step is required to finish problem, and that is to factor thex2term and

the32term intox∗xand3∗3. If desired, instructor mentions property of exponents,

(ab )2=a2b2

(x+3)(x∗x−3 x+3∗3)

( x+3 )(x2−3 x+9)

Instructor needs to be cognizant some students need to see each step, so that is

why decompose the last trinomial and show how to multiplyx∗xand3∗3.

Example 2: 27m3+64q3



numbers 27∧64 are considered perfect cubes. This leads to the assertion that a sum of

cubes is present. Advise the students to decompose, or factor each term separately.

Proceed as follows:

69

3m∗3m∗3m+4q∗4 q∗4 q


(3m )3+ (4 q )3


are.

first term=3m

last term=4 q

Following the format of the sum of cubes formula, instructor advises to insert

terms. For the benefit of the students, write down formula on board:

(F+ L)(F2−FL+L2)

(3m+4 q )((3m )2−12mq+(4 q )2)

One more step is required to finish problem, and that is to factor the(3m)2term

and the(4q)2term into3m∗3mand4 q∗4 q. If desired, instructor mentions property of

exponents,(ab )2=a2b2

(3m+4 q)(3m∗3m−12mq+4 q∗4 q)

(3m+4 q )(9m2−12mq+16q2)

Instructor needs to be cognizant some students need to see each step, so that is why

decompose the last trinomial and show how to multiply3m∗3mand4 q∗4 q.

Closure (Reflect Anticipatory Set): Use polynomial division (long or synthetic) to

determine if one polynomial is a factor of another. If time permits, show how

factoring formula can be found by long division.

Plan for Independent Practice: Worksheet contained in student packet provided by

workshop.

70

Sample problems for instructor use, if so desired:

Factor completely:

1) y3+z3

2) h3+64

3) y6+125

4) 216d3+343

5) 64h3+b9

6) Fill in the missing parts: x3+ y3=¿

7) Fill in the missing parts: 64 r 6+¿___¿(___+¿___)(___−4 r2 p+ p2 ¿

8) Fill in the missing parts: 125 g3+64m3=¿(___+¿___¿¿___+¿___)

9) Fill in the missing parts: ___+¿___¿(d+5 z )(d2−5dz+25 z2)

10) Fill in the missing parts: ___+¿___¿(4 r+5 k2)¿ ___−¿___+¿___)

Answer key to sum of cubes sample problems:

1) y3+z3=( y+ z)( y¿¿2− yz+ z2)¿

2) h3+64=(h+4)(h2−4h+16)

3) y6+125=( y2+5)( y4−5 y2+25)

4) 216d3+343=(6d+7)(36 d2−42d+49)

5) 64h3+b9=(4 h+b3)(16h2−4b3h+b6)

6) x3+ y3=( x+ y )(x2−xy+ y2)

7) 64 r 6+ p3=(4 r2+ p)(16 r4−4 r2 p+ p2)

8) 125 g3+64m3=(5g+4m)(25 g2−20 gm+16m2)

9) d3+125 z3=(d+5 z )(d2−5dz+25 z2)

71

10) 64 r 3+125k 6=(4 r+5 k2)(16 r2−20k2r+25k4)

References


Boston: Pearson.


July 30, 2010 from


12.htm






Company.



End of Lesson Plan

72

http://www.lessonplanspage.com/MathFactoringPolynomialsSquaresAndCubes912.htm

http://www.lessonplanspage.com/MathFactoringPolynomialsSquaresAndCubes912.htm

Lesson Plan 7: Difference of Cubes

Instructor Notes

73

Lesson Plan Title: Difference of Cubes


specifically two term expressions, which are called binomials.








and algorithms to combine, transform, and evaluate expressions (e.g., polynomials,




and algorithms to solve equations, inequalities, and systems of linear equations.



D. Representations



relationships.

General Goal(s): Identify and factor completely polynomials that are difference of

cubes.

74

Specific Objectives: Perform operations with polynomials, specifically two term

expressions, which are called binomials.

Required Materials:





Anticipatory Set (Lead-In): Instructor again would lead into discussion of perfect cube

numbers with the students. Instructor refers students to cubed numbers in chart format

provided in student packet.


Common methods of factoring polynomials are based on the number of its terms:

1) For any number of terms, factor out the greatest common factor(s) first.

2) For four or more terms, group like terms and factor out common factors.

3) For two terms:

Factor the sum of two cubes into the product of a binomial sum and

quadratic trinomial whose second term is negative.

Factor the difference of two cubes into the product of a difference binomial

and quadratic trinomial with all positive terms.

Difference of Cubes

The formula for the difference of two cubes is: A3−B3=(A−B)(A2+AB+B2)

A=first term∧B=last term

75

Some teachers prefer the following format:

F3−L3=(F−L)(F2+FL+L2)


At this time, instructor discusses with students the importance of recognizing

perfect cube numbers, variables, or complete expressions. Some examples are

27 , x9 , ( x+7 )6.

Example 1: x3−27

For this type problem, two terms are present, which implies a binomial, a minus


number 27 is considered a perfect cube. This leads to the assertion that a difference of

cubes is present. Advise the students to decompose, or factor each term separately.

Proceed as follows:

x∗x∗x−3∗3∗3


( x )3−(3 )3


are.

first term=x

last term=3

Following the format of the difference of cubes formula, instructor advises to

insert terms. For the benefit of the students, recopy formula on board:

(F−L)(F2+FL+L2)

76

( x−3 )( ( x )2+3 x+(3 )2)

One more step is required to finish problem, and that is to factor the(x)2term and

the(3)2term intox∗xand3∗3. If desired, instructor mentions property of exponents,

(ab )2=a2b2

(x−3)(x∗x+3 x+3∗3)

( x−3 )( x2+3 x+9)


why decompose the last trinomial and show how to multiplyx∗xand3∗3.

Example 2: 27m3−64 q3

For this type problem, two terms are present, which implies a binomial, a minus


numbers 27∧64 are considered perfect cubes. This leads to the assertion that a

difference of cubes is present. Advise the students to decompose, or factor each term

separately. Proceed as follows:

3m∗3m∗3m−4q∗4q∗4q


(3m )3−( 4q )3


are.

first term=3m

last term=4 q

77

Following the format of the difference of cubes formula, instructor advises to

insert terms. For benefit of students, write down formula on board:

(F−L)(F2+FL+L2)

(3m−4q )((3m)2+12mq+(4 q )2)

One more step is required to finish problem, and that is to factor the(3m)2term

and the(4q)2term into3m∗3mand4 q∗4 q. If desired, instructor mentions property of

exponents,(ab )2=a2b2

(3m−4q)(3m∗3m+12mq+4 q∗4 q)

(3m−4q )(9m2+12mq+16q2)


why decompose the last trinomial and show how to multiply3m∗3mand4 q∗4 q.

Closure (Reflect Anticipatory Set): Use polynomial division (long or synthetic) to

determine if one polynomial is a factor of another. If time permits, show how

factoring formula can be found by long division.

Plan for Independent Practice: Worksheet contained in packet provided at workshop.


1) c3−64

2) x3−1

3) y3−8

4) p3−27

5) 8−27b3

6) 64−125 x3

78

7) a3−b3

8) x3− y3

9) y3−128

10) 3 z3−3

Answer key to difference of cubes sample problems:

1) (c−4)(c2+4 c+16)

2) (x−1)(x¿¿2+ x+1)¿

3) ( y−2)( y2+2 y+4)

4) ( p−3)( p¿¿2+3 p+9)¿

5) (2−3b)(4+6b+9b¿¿2)¿

6) (4−5 x )(16+20 x+25 x¿¿2)¿

7) (a−b)(a2+ab+b2)

8) (x− y )(x¿¿2+xy+ y2)¿

9) 2( y−4)( y¿¿2+4 y+16)¿

10) 3(z−1)(z¿¿2+z+1)¿

References


Boston: Pearson.

Bouyssounouse, J. (2010). Sum of two cubes worksheet. Edonyourown.com. Retrieved

August 1, 2010 from http://www.edonyourown.com/




79

http://www.edonyourown.com/



Company.



End of Lesson Plan

80

Lesson Plan 8: Monic Trinomials

Instructor Notes

81

Lesson Plan Title: Factoring Monic Trinomials


specifically three term expressions, which are called trinomials.





















82



D. Representations



General Goal(s): Ability to factor monic trinomials efficiently, recognize their presence,

and become proficient in determining factors of large numbers. Monic quadratic

trinomials are expressions where the leading coefficient is equal to 1.

Required Materials:





Anticipatory Set (Lead-In): Discuss with students that workshop deals with exploring

all different types of factoring scenarios. Trinomials are only one type of polynomial that

the student will be exposed to while in college.


Quadratic trinomials have the forma x2+bx+c. When factoring these trinomials,

we want to write them as a product of two binomials, if they are indeed factorable.

Monic quadratic trinomials are expressions where the leading coefficient is equal

to 1. In other words, the number in front of the x2 term is a 1. For example,x2+17 x+66 is

a monic quadratic trinomial. These monic trinomials are the simplest to factor. The

overall strategy when dealing with a monic trinomial is the following:

83

1) Focus on the last term in the trinomial. The instructor and students need to

find paired factors of the last term, that is, what two unique numbers when multiplied

together will produce the constant in the trinomial. The constant is the number in the

trinomial that has no variable attached to it.

2) The next step is to think of what two paired factors that when multiplied

together produced the constant, but also those two unique paired factors if added or

subtracted will also produce the coefficient in front of the middle term.

3) Instructor advises students to not worry about signs in the original trinomial.

The signs will “fall into place”, so to speak at a later time.

4) If the instructor or students find paired factors of the last term in the trinomial

that if added or subtracted will produce the coefficient in front of the middle term, then

the monic trinomial is indeed factorable.

Example: x2+26 x+88

1) The instructor and students together need to find all paired factors of 88 that if

the sum or difference of those two unique paired factors equals2 6, then the trinomial is

indeed factorable. Instructor advises students to list all paired factors; since it is good

practice for the remedial level students. Mention to students that organization is needed

in the factoring process. Proceed as follows:

Paired Factors of 88 Sum Difference

88*1 88+1=89 88-1=87

44*2 44+2=46 44-2=42

22*4 22+4=26 22-4=18

84

11*8 11+8=19 11-8=3

2) Instructor ask the students what two unique paired factors multiplied together

will produce the last term, but also if the student adds or subtracts the two unique paired

factors, the coefficient of the middle term is also produced.

3) Instructor and student will try all combinations of factors, which may take some

time for the beginning student.

4) There are only two unique paired factors that when multiplied together will

produce the last term, and also if added or subtracted will produce the coefficient of the

middle term. Those two unique factors are22∧4.

5) At this time, the instructor and student know the trinomial is factorable and can

be written as a product of two binomials.

6) The instructor will discuss the importance of being organized while performing

the factoring steps. If so desired, the instructor can show step-by-step the correct

procedure to factor the trinomial. After a few more problems, some steps can be

consolidated into one smooth sequence. This step-by-step process is for benefit of

students being first exposed to the factoring techniques. Mention to students that

organization is needed in the factoring process. Proceed as follows by listing problem

and justification as a heading.


7) x2+26 x+88 Recopy problem.

8) ()() “Open close, open close”.

9) (x)( x) Insert variable.

10) (x 4)(x 22) Insert numbers.

85

For this type of problem, the ordering of numbers being inserted does not matter.

The instructor needs to discuss with students that if they were to place the number 22

first, then the number 4 second, it would not change anything in the problem. State if

signs from original problem contain two “plus” signs, then two “plus signs” are inserted

as last step in factoring process.

11) (x+4 )( x+22) Final answer

Example: y2−10 y−39

1) The instructor and students need to find paired factors of 39 that if the sum or

difference of those two unique paired factors equals10, then the trinomial is indeed

factorable. Discuss with students not to be concerned with sign in front of the middle

term. The signs will fall into place at a later time. Instructor advises students to list all

paired factors; since it is good practice for the remedial level students. Proceed as

follows:


39*1 39+1=40 39-1=38

13*3 13+3=16 13-3=10

2) Instructor asks the students what two unique paired factors multiplied together





86



middle term. Those two unique factors are13∧3.

5) At this time, the instructor and student indeed know the trinomial is factorable

and can be written as a product of two binomials. The instructor will discuss the

importance of being organized while performing the factoring steps. If so desired, the

instructor can show step-by-step the correct procedure to factor the trinomial. After a

few more problems, some steps can be consolidated into one smooth sequence. This

step-by-step process is for benefit of students being first exposed to the factoring

techniques. Mention to students that organization is needed in the factoring process.

Proceed as follows by listing problem and justification as a heading.


6) y2−10 y−39 Recopy problem.


8) ( y )( y ) Insert variables.

9) ( y 3)( y13) Insert numbers.


The instructor needs to discuss with students that if they were to place the number13

first, then the number 3 second, it would not change anything in the problem.

Insert signs according to the sign that precedes the middle term, from the original

problem! The sign that falls in front of middle term, that is, the 10 y term has a negative

sign in front of it; therefore, the negative sign is positioned in front of the larger of the

two numbers. We know 13 apples are much larger than 3 apples, so the negative sign

87

goes in front of the13. Advise students to insert numbers first, then signs. If both signs

from original problem are negative, then student will have both “positive and negative

signs” in final answer.

10) ( y−13)( y+3) Final answer

Example: x2−27 x+72

1) The instructor and students need to find all paired factors of 72 that if the sum

or difference of those two unique paired factors equals27, then the trinomial is indeed

factorable. Instructor advises students to list all paired factors; since it is good practice

for the remedial level students. Proceed as follows:


72*1 72+1=73 72-1=71

36*2 36+2=38 36-2=34

24*3 24+3=27 24-3=21

18*4 18+4=22 18-4=14

12*6 12+6=18 12-6=6

9*8 9+8=17 9-8=1






88



middle term. Those two unique paired factors are24∧3.

5) At this time, the instructor and student indeed know the trinomial is factorable

and can be written as a product of two binomials. The instructor will discuss the

importance of being organized while performing the factoring steps. If so desired, the

instructor can show step-by-step the correct procedure to factor the trinomial. After a

few more problems, some steps can be consolidated into one smooth sequence. This

step-by-step process is for benefit of students being first exposed to the factoring

techniques. Be sure and reiterate from time to time to not be overly concerned about

signs in original problem. Signs will fall into place at a later time. Proceed as follows by

listing problem and justification as a heading.


6) x2−27 x+72 Recopy problem.


8) (x)( x) Insert variables.

9) (x 24)(x3) Insert numbers.


The instructor needs to discuss with students that if they were to place the number3first,

then the number 2 4 second, it would not change anything in the problem.

If last term is positive, and middle term negative from original problem, then both

signs will be negative in final answer.

10) (x−24 )( x−3) Final answer

89

The instructor must emphasize the importance of inserting the numbers first, then

signs. This trick helps reduce stress levels for students dramatically. The overall

strategy for monic trinomials is to look at last term and find paired factors of that unique

number that might add or subtract to equal the middle term’s coefficient. Stress to the

students that the signs will “fall into place”, so to speak, at a later time.


Worksheet contained in packet provided at workshop.


Factor the following trinomials.

1) x2+7x+12

2) x2+4 x−12

3) x2−9 x+8

4) x2+11x−26

5) x2+16x+55

6) x2−3 x−10

7) x2+113 x+222

8) x2−8 x−20

9) x2+105 x+404

10) x2+11x+30

Answer key to monic trinomial sample problems:

1) (x+4 )( x+3)

2) (x+6)(x−2)

3) (x−1)(x−8)

90

4) (x+13)(x−2)

5) (x+11)(x+5)

6) (x−5)(x+2)

7) (x+111)(x+2)

8) (x−10)(x+2)

9) (x+101)(x+4)

10) (x+5)(x+6)

References


Boston: Pearson.




Karassev, A. (2010). Polynomial tutorial. Retrieved August 1, 2010 from Nipissing

University: http://www.nipissingu.ca/



Company.



End of Lesson Plan

91

http://www.nipissingu.ca/

Lesson Plan 9: Non-Monic Trinomials

Instructor Notes

92

Lesson Plan Title: Factoring Non-Monic Trinomials


specifically three term expressions, which are called trinomials.





















93



D. Representations



relationships.

General Goal(s): Being able to factor non-monic trinomials efficiently and determine

their presence by inspection only. Non-Monic trinomials have a leading coefficient of

not equal to 1.

Required Materials:





Anticipatory Set (Lead-In): Discuss with students that workshop deals with exploring

all different types of factoring scenarios that the students may see in their classes. Non-

Monic Trinomials are only one type that the student will be exposed to while in college.

Step-By-Step Procedures:Quadratic non-monic trinomials have the forma x2+bx+c. When factoring these

trinomials, we want to write them as a product of two binomials.

Non-monic quadratic trinomials are expressions where the leading coefficient a is

not equal to 1. For example 70 x2−51x−70 is a non-monic quadratic trinomial. The

overall strategy to factor this type of trinomial is to find paired factors of a∗c whose sum

or difference equalsb, then the trinomial is indeed factorable.

94

Example: 13 x2+24 x+11

A*C Method

a x2+bx+c

1) We will use the a∗cmethod, which is multiplying the first coefficient times the

constant. For this problem, multiply13∗11=143. We need to find all paired factors of143,

whose sum or difference equals the middle coefficient. The instructor and student will

multiply the first term’s coefficient by the last term, which is the constant. Do not be

overly concerned with the signs; just take13∗11. List all pairs of integers with product143

, which will best be put into some form of a table so students can see numbers more

clearly. Proceed as follows:

Paired factors Sum Difference

13*11 13+11=24 13-11=2

143*1 143+1=144 143-1=142

2) The instructor is only looking for paired factors of 143 if added or subtracted

might equal the middle term’s coefficient or for this example the number24, then the

trinomial is considered factorable. At this time, the sum of the two paired factors does

indeed equal the middle coefficient; therefore, this non-monic trinomial can now be

broken down into a four term polynomial by decomposing the middle term only into two

terms. Proceed as follows by listing problem and justification as a heading.


3) 13 x2+24 x+11 Recopy problem.

95

4) 13 x2+13 x+11 x+11 Decompose only the middle term.

Instructor can use different colored pens for more distinction.

5) (13 x2+13 x )+(11 x+11) Group first two terms, then group last

two terms.

6) (13 x∗x+13x∗1 )+(11∗x+11∗1) Decompose, or factor each term.

7) 13 x ( x+1 )+11(x+1) Factor GCF’s.

8) (x+1)(13 x+11) Insert common factor first,

followed by monomial factors into

last set of parenthesis.

9) (x+1)(13 x+11) Final answer.

A*C Method

a x2+bx+c

Example: 2 x2+9 x+7



whose sum or difference equals the middle coefficient. The instructor and students will


overly concerned with the signs; just take2∗7. List all pairs of integers with product14 ,

which will best be put into some form of table, so students can see numbers more


96

Paired Factors Sum Difference

14*1 14+1=15 14-1=13

7*2 7+2=9 7-2=5



trinomial is considered factorable. At this time, the sum of the two paired factors does

indeed equal the middle coefficient; therefore, this non-monic trinomial can now be




3) 2 x2+9 x+7 Recopy problem.

4) 2 x2+2 x+7x+7 Decompose only the middle term.


5) (2 x2+2 x )+(7 x+7) Group first two terms, then group last

two terms.

6) (2 x∗x+2x∗1 )+(7∗x+7∗1) Decompose, or factor each term.

7) 2 x ( x+1 )+7(x+1) Factor GCF’s.

8) 2 x ( x+1 )+7(x+1)

97

9) (x+1)(2x+7) Insert common factor first,


last set of parenthesis.

10) (x+1)(2x+7) Final answer.

How does a student check his or her work in the factoring process? Just do

FOIL, which stands for:

First

Outer

Inner

Last

A*C Method

a x2+bx+c

Example: 2 x2−x−28



whose sum or difference equals the middle coefficient. The instructor and students will


overly concerned with the signs; just take2∗28. List all pairs of integers with product56 ,

which will best be put into some form of table, so students can see numbers more


98

Paired Factors Sum Difference

56*1 56+1=57 56-1=55

28*2 28+2=30 28-2=26

14*4 14+4=18 14-4=10

8*7 8+7=15 8-7=1



trinomial is considered factorable. At this time, the difference of the two paired factors

does indeed equal the middle coefficient; therefore, this non-monic trinomial can now be




3) 2 x2−x−28 Recopy problem.

4) 2 x2+8 x−7 x−28 Decompose only the middle term.

5) (2 x2+8x )+(−7 x−28) group accordingly, being sure to

place plus sign between the

parentheses to avoid confusion

with students.


6) (2 x∗x+2 x∗4 )+(−7∗x−7∗4 ) Decompose, or factor each term.

99

7) 2 x ( x+4 )−7 (x+4) Factor common terms, being sure

to change the sign of last term

only. The last term is the4.

8) (x+4 )(2x−7) Insert common factor first,


last set of parentheses.

9) (x+4 )(2x−7) Final answer.

Example: 35 x2−42 x−77

Here is the huge drawback to thea∗cmethod, and that is the student will take two

huge numbers, multiply them together and produce a much bigger number. This creates

stress for the beginning students in the factoring segment of their respective classes.

One reason for this is that a great deal of students do not know how to “break down”

large integers into prime factors. The situation only gets more complicated when finding

factors of the “much bigger number”. One way to avoid this situation is to do the

following:

1) Decompose the first term into smaller paired factors immediately, do same for

the last term, which is the constant. This keeps numbers more manageable from the

start and mental computation can be performed easier.

2) The instructor will take two products and either adds or subtracts the two

products and if the middle term is attained or reached, then this trinomial is indeed

factorable. Proceed as follows:

100

Paired Factors of first term Factors of last term

35x 77

x 1

What other numbers multiplied together will produce35 x2or77? Instructor stresses

to students to find ALL paired factors of first and last term. This may take some time for

beginning students, but after practice, it will become easier. See other paired factors

below:

7x 11

5x 7

3) The instructor will take initially the 35 x∧x paired factors and the77∧1 paired

factors and determines if two products by adding or subtracting will yield42 x. The

instructor multiplies straight across or does the criss-cross pattern, if so needed. It is

much easier to see visually, therefore proceed as follows:

35x 77

x 1

Product

35 x∗77=2695 x (First product)

x∗1=x (Second product)

4) At this time, the instructor either adds or subtracts the first product and the

second product, and if 42 x is achieved, then the trinomial is indeed factorable.

Sum

2695 x+ x=2696 x

101

Difference

2695 x−x=2694 x

5) The middle term of 42 x has not been attained at the present time. The

instructor will attempt again using the criss-cross pattern, which is allowed.

35x 77

x 1

Product

35 x∗1=35 x(First product)

x∗77=77 x(Second product)

At this time, the instructor either adds or subtracts the first product and the

second product, and if 42 x is achieved, then the trinomial is indeed factorable.

Sum

35 x+77x=112x

Difference

77 x−35 x=42x

The middle term of 42 x has now been attained. The non-monic trinomial is

indeed factorable. Instructor advises students to make arrows to keep track of pairing of

numbers.

35x 77

x 1

102

At this time, instructor will discuss process with students of multiplying straight

across or criss-cross, if needed. For the above problem, instructor tried different

combinations of the first paired factors, which was35 x2with the paired factors of the last

term, which was77. It just so happened that after two steps in the process, the middle

term of42 x was reached. This might not always be the case, though. The instructor will

continue to try different combinations of paired factors until the middle term has been

reached. This may take several attempts, especially if first coefficient and last term are

large integers.

At this time, instructor will proceed to complete factoring process. Proceed as

follows by listing problem and justification as a heading.


6) 35 x2−42 x−77 Recopy problem.

7) ()() “Open, close, open, close”.

8) (35 x)(x ) Insert factors from first column.

9) (35 x)(x 1) The 35 x∧1 represent the OUTER portion of

FOIL process. That is why arrows are used.

10) (35 x77)(x 1) The x∧77 represent the INNER portion of FOIL

process. That is why arrows are used.

Signs are placed last in process. The question that is always asked is “Where do

the signs go, and why”? At this time, the instructor will perform only the OUTER, then

INNER part of the FOIL process. Proceed as follows:

OUTER: 35 x

INNER: 77 x

103

Difference:35 x−77 x=−42x

In order to ensure the correct middle term’s coefficient, the student must subtract

the OUTER and INNER, but the negative sign must go in front of the 77 to be sure

−42x is obtained when subtracting.

11) (35 x−77)(x+1)

12) (35 x−77)(x+1) Final answer.

This technique is one of several methods to factor the non-monic trinomials. The

arrows used help with order of multiplication. One trick to tell students is to use

Popsicle® sticks for the “arrows”. The sticks are cheap to purchase, and easy to use.


Worksheet contained in packet provided by workshop.


1) 2 x2+15 x+28

2) 35 g2+99 g−72

3) 10 x2−7 x−12

4) 33 z2−4 z−37

5) 8a2−26a+15

6) 15q2+56q+49

7) 2 x2+x−6

8) 15b2−14 b−49

9)3 x2−11 x+6

10) 19h2−80h−99

104

Answer key to non-monic trinomial sample problems:

1) (2 x+7 ) ( x+4 )

2) (7 g+24 )(5g−3)

3) (5 x+4 ) (2 x−3 )

4) (33 z−37 )( z+1)

5) (4 a−3 ) (2a−5 )

6) (5q+7 )(3q+7)

7) (2 x−3 ) ( x+2 )

8) (5b+7 ) (3b−7 )

9) (3 x−2 ) ( x−3 )

10) (19h−99 ) (h+1 )

References


Boston: Pearson.








Company.

105



End of Lesson Plan

106

Lesson Plan 10: Techniques of Factoring

Instructor Notes

107

Lesson Plan Title: Techniques of factoring

Student Learning Objective: Determine best method to solve various quadratic,

rational or word problems.














General Goal(s): Student’s exposure to solving various quadratic word problems and

the ability to determine best method to solve.

Specific Objectives: Given a certain type of quadratic equation, students can

determine best method to solve. The following techniques can be employed:

determining greatest common factor, grouping method, sum or difference of squares,

sum or difference of cubes, monic and non-monic trinomials, and if working with rational

type expressions, determining the least common denominator (LCD).

108

Required Materials:





Anticipatory Set (Lead-In): Discuss with students at this time the problems and

techniques that have been shown so far will come into play now that students will be

required to solve various quadratic equations, rational expressions or word problems

that containa x2+bx+c.


Instructor advises students to:

1) Read problem very carefully several times.

2) Determine what information is given.

3) Recognize the type of problem.

4) The problem they are reading may be asking to determine a person’s age,

finding distance, interest on a loan, or some type of work problem.

Word Problems

5) Assign variables that are appropriate to the problem. For example, uset for

time, r for rate, d for distance, lfor length, and w for width, to name a few.

6) Translate the mathematical language into English language. Advise students

this is the hardest part, but will become easier with practice.

109

7) Solve for the variable using any method or technique discussed in workshop

so far. Remind students workshop is to reinforce concepts that have been already

covered in classroom.

8) Check answers to be sure they make sense in context of original problem. For

example, a student cannot have negative time or a person’s age cannot be negative.


Worksheet contained in packet provided at workshop.

References


Boston: Pearson.

Duval, R. Useful tips to solving algebra word problems. (2008). www.articlesbase.com.

Retrieved August 1, 2010 from http://www.articlesbase.com/






Company.



End of Lesson Plan

110

Lesson Plan 11: Solving an Equation for a Certain Variable

Instructor Notes

111

Lesson Plan Title: Solving a Quadratic or Rational Equation for a Specified Variable

Student Learning Objective: Given quadratic or rational equations, ability to solve for

a certain variable.





















112



D. Representations



relationships.

General Goal(s): Students in elementary and intermediate level will be able to factor,

and use zero product property to determine answer. College algebra students will use

previous techniques discussed, but also solve using the quadratic formula, and by

completing the square.

Required Materials:






Definition: A quadratic equation is of the forma x2+bx+c=0, wherea ,b ,∧care real

numbers, anda≠0. (Swokowski, 1993)

Instructor advises students it is critical to recognize types of quadratic equations.

To determine if a quadratic equation is present, look for the following:

1) Ifa≠0, then the equation must have a variable raised to the second power.

113

2) Ifb=0, then the equation simplifies toa x2+c=0, which is still considered

quadratic.

3) Ifc=0, then the equation simplifies toa x2+bx=0, which is still considered

quadratic.

4) Ifb∧c=0 then equation simplifies toa x2=0, which is still considered quadratic.

5) Ifa=0, then equation simplifiesbx+c=0, which is linear.

Solving a Quadratic Equation if b=0

The following technique of using the square root property applies only to

intermediate and college algebra students.

Example: y=5 x2−20

Instructor will advise students to “set equation equal to 0”, and proceed as

follows:


5 x2−20=0

5 x2−20+20=0+20 Add 20 to both sides of equation.

5 x2=20

5x2

5=20

5Divide by 5 to both sides of equation.

x2=4

At this time, instructor discusses with students the concept of "doing and

undoing" processes in mathematics. In order to undo a square power, taking the square

root of both sides of equation is necessary. Proceed as follows:

114

√ x2=±√4 Advise students to “take square root” on

both sides of equation, being sure to

insert ±sign on right side of

equation.

x=±√4

x=±2

x=2∨x=−2

Instructor will ask students what other method could have been used to solve the

previous problem. Try and solve using previous topics covered in workshop. Proceed as

follows:


1) y=5 x2−20 Recopy equation.

2) 5 x2−20=0 Set equation equal to zero.

Instructor asks students if a common factor is present. Both terms do share a

factor of5. Proceed:

3) 5x2

5−20

5=0

5Divide by5.

4) x2−4=0

At this time, a binomial is present on left side of equation represented by a

difference of squares. This can be factored accordingly. Proceed:

5) ( x−2 ) ( x+2 )=0

At this time, advise students to set each factor equal to zero and solve forx.

6) x−2=0∨x+2=0

115

7) x−2+2=0+2 x+2−2=0−2

8) x=2 x=−2

9) x=±2

Solving a Quadratic Equation if c=0

Now, supposec=0 , then we havey=a x2+bx. Setting y=0 we have:

a x2+bx=0

Factor out anx:

x (ax+b )=0

Now we have two factors whose product is 0. One or the other must be zero;

otherwise their product is not zero, according to the zero product property.

Zero-Product Property

Ifab=0 , then a=0∨b=0 ,∨both (Sullivan, 2002).

Either x=0∨ax+b=0

Solving for x we have x=0∨x=−ba

So, the solution isx=0∨x=−ba

Solving a Quadratic Equation if a ,b , care Non Zero

116

Considering the case wherea ,b ,and c are all nonzero, we can use the quadratic

formula or attempt to factor a monic or non-monic trinomial.

College Algebra Students Only (if time permits)

For the students in college algebra, proceed as follows using the formula:

x=−b±√b2−4ac2a

Example: y=x2+8 x+7

It is important to set equation equal to zero before determining the coefficients.

Proceed as follows:

1) x2+8 x+7=0

At this time, instructor advises students to determine whata ,b ,∧c are.

a=1b=8c=7

Using the quadratic formula, the instructor advises the students to insert the

numbers correctly into the formula.

x=−8±√82−4∗1∗72∗1

x=−8±√64−282

x=−8±√362

x=−8±62

x=−8−62

∨x=−8+62

x=−142

∨x=−22

117

x=−7∨x=−1

Instructor advises students that the previous problem could also have been

solved by the monic trinomial method.

1) x2+8 x+7=0








middle term. Those two unique paired factors are7∧1.

5) At this time, the instructor and student know the trinomial is factorable and can

be written as a product of two binomials.

6) The instructor will discuss the importance of being organized while performing

the factoring steps. If so desired, the instructor can show step-by-step the correct

procedure to factor the trinomial. After a few more problems, some steps can be

consolidated into one smooth sequence. This step by step process is for benefit of

students being first exposed to the factoring techniques. Proceed as follows:


7) x2+8 x+7=0 Recopy problem.

8) ()()=0 “Open close, open close”.

9) ( x ) (x )=0 Insert variable.

118

10) ( x7 ) ( x 1 )=0 Insert numbers.


The instructor needs to discuss with students that if they were to place the number 1

first, then the number 7 second, it would not change anything in the problem. State if

both signs from original problem contain two “plus” signs, then two “plus signs” are

inserted as last step in factoring process.

11) ( x+7 ) ( x+1 )=0

12) x+7=0∨x+1=0 Set each factor equal to zero.

13) x+7−7=0−7 x+1−1=0−1

14) x=−7 x=−1

15) x=−7∨x=−1

Advise students that there are different ways to solve quadratic equations. Allow

students some creativity in thinking of what methods could be used.

College Algebra Students Only

The Discriminant

In the quadratic formula, the portion √b2−4ac is called the discriminant. If this

discriminant is equal to zero then there is only one solution to the quadratic equation,

namelyx=−b2a .

If this discriminant is negative then there are no real solutions to the quadratic

equation, which implies to college algebra students, imaginary numbers using i are

119

present. Otherwise there are two solutions to the equation, leading to two unequal real

solutions.

Values of Discriminantb2−4 ac

Nature of the Roots of : a x2+bx+c=0

Positive value 0 Negative value

Two real and unequal roots One root of multiplicity two No real root

(Swokowski, 1993)

Derivation of the quadratic formula

The quadratic formula comes from a technique called completing the square.

a x2+bx+c=0(Equation 1)

Our goal is to rewrite the equation so that we can factor it into something similar

to( x+d )2.

In order to perform the process of completing the square, the coefficient in front

of thex2 term must be equal to1. Therefore, divide bya. Proceed as follows:

a x2+bx+c=0

a x2

a+ bxa

+ ca=0a

Notice how theacoefficient in the first term cancels out, leaving only thex2term.

x2+ bxa

+ ca=0

At this point, instructor will discuss the process of “completing the square” to the

students. Proceed as follows by subtracting ca from both sides.

120

x2+ bxa

+ ca− ca=0− c

a

x2+ bxa

=−ca

At this point, the instructor will advise students to take 12of

ba term then square

the results.

12∗b

a= b

2a

( b2a )2

= b2

4 a2

Instructor will recopy equation 2, and then add b2

4 a2 term to both sides of

equation.

x2+ bxa

=−ca

x2+ bxa

+ b2

4a2 =−ca

+ b2

4a2

Rearranging terms on right side of equation, instructor will proceed:

x2+ bxa

+ b2

4a2 =b2

4 a2 −ca

At this time, the left side of equation is a trinomial that needs to be rewritten into

factored binomial form. The process involved is to enclose within parenthesis thexterm

only followed by plus sign, thenb

2a term being sure the grouping is squared.

(x+ b2a )

2

= b2

4 a2 −ca

121

The concern for right side of equation is that instructor has two terms with

different denominators, so therefore, a single quotient is needed. Noting the LCD is4 a2 ,

the instructor performs the following:

(x+ b2a )

2

= b2

4 a2 −

ca∗4a

4a

(x+ b2a )

2

= b2

4 a2 −4ac4 a2

At this time, the instructor will discuss the importance of being able to undo a

squaring operation, by “taking” the square root of both sides of equation. Stress to

students to insert ± sign on right side of equation only. Proceed as follows:

√(x+ b2a )

2

=±√ b2−4ac4a2

x+ b2a

=± √b2−4ac√4a2

x+ b2a

=± √b2−4ac2a

A few more steps are involved for completion of derivation of quadratic formula,

and that is to move b

2a term to right side of equation. Proceed as follows by subtracting:

x+ b2a

− b2a

=−b2a

± √b2−4ac2a

x=−b2a

± √b2−4ac2a

The final step is to combine the two fractions on right side of equation into a

single denominator.

122

x=−b±√b2−4ac2a

Solving a Quadratic Equation

An equation in which the variable is raised to the second power or has the form

a x2+bx+c=0 is considered a quadratic equation. Before solving quadratic equations by

the method of factoring, it is necessary to remember the zero product property.


Ifab=0 , then a=0∨b=0 ,∨both (Sullivan, 2002)

Solve the following:

Example 1: x2+5x=0

1) x (x+5 )=0 Factor out GCF.

2) x=0 ¿ x+5=0 Set each factor equal to zero.

3) x=0 ¿ x+5−5=0−5 Subtract 5 from both sides.

4) x=0 ¿ x=−5

Example 2: x2−x=6

Notice that this equation does not equal zero. The first step is to subtract 6 from

each side.

1) x2−x−6=6−6

2) x2−x−6=0

123

At this time, instructor will remind students that a monic trinomial is present.

Factor accordingly from skills learned in monic trinomial section.

3) ( x−3 ) ( x+2 )=0

4) x−3=0 ¿ x+2=0

5) x−3+3=0+3 x+2−2=0−2

6) x=3 ¿ x=−2

Solving Rational Equations

Example: 2c15

−23=1

3− c

5

The most important strategy for rational expressions that can be performed is to

“clear fractions” by multiplying each term by the LCD (least common denominator). In

one swift motion, so to speak, all denominators are cancelled leaving only the

numerators. Proceed as follows:


1) 2c15

−23=1

3− c

5 Recopy equation, noting LCD is 15.

2) 15∗( 2c15

−23)=15(¿ 1

3− c

5) Multiply both sides of equation by

LCD.

3) 15∗2c

15−15∗2

3=15∗1

3−15∗c

5 Distribute “15” to each term.

4) 15∗2c

15−5∗3∗2

3=5∗3∗1

3−5∗3∗c

5 Decompose each term.

124

At this point, to see more clearly the steps involved in reducing fractions,

instructor tells students to make it a point to use a different colored pen or pencil and

cross out the terms that will remove a factor of 1. For this application, the red colored

numbers will remove a factor of 1, leaving only the numerators in the expression.

5) 15∗2c

15−5∗3∗2

3=5∗3∗1

3−5∗3∗c

5

6) 2c−5∗2=5∗1−3∗c "Clearing fractions".

7) 2c−10=5−3c Intermediate result.

8) 2c−10+3c=5−3c+3c Add3c to each side.

9) 5c−10=5

10) 5c−10+10=5+10 Add 10 to each side.

11) 5c=15

12) 5c5

=155 Divide each side by 5.

13) c=3 Final answer

The most important step was to multiply each term by the LCD, which “cleared

out” fractions efficiently. This critical step will be performed for rational expressions in

order to make further computations and manipulations easier.

Example:x+5

2x2−3 x−2+ x−1x2−5 x+6

= 3 x+42x2−5 x−3

Dealing with a rational expression, it is very important to factor only the

denominators at this time. The denominators are monic and non-monic trinomials,

which were discussed from previous sections. Proceed as follows:


125

1) x+5

(2x+1 ) ( x−2 )+ x−1

( x−2 ) ( x−3 )= 3 x+4

(2x+1 ) ( x−3 ) Factor the denominators.

Note LCD is (2 x+1)(x−2)(x−3)

At this point, make it a point to realize that to maintain balance in the entire

equation; the LCD must be multiplied to ALL terms!

2) (2 x+1 ) ( x−2 ) ( x−3 )∗[ x+5(2 x+1 ) ( x−2 )

+ x−1( x−2 ) (x−3 )

= 3 x+4(2 x+1 ) (x−3 )

] Multiply entire

expression by LCD.

3) (2 x+1 ) ( x−2 ) ( x−3 )∗x+5(2 x+1 ) ( x−2 )

+(2 x+1 ) ( x−2 ) (x−3 )∗x−1( x−2 ) ( x−3 )

=(2 x+1 ) ( x−2 ) ( x−3 )∗3x+4(2x+1 ) ( x−3 )

To see more clearly the steps involved in reducing fractions, instructor tells

students to make it a point to use a different colored pen or pencil and cross out the

terms that will remove a factor of 1. For this application, the red colored expressions will

remove a factor of 1, leaving only the numerators in the expression.

4)(2 x+1 ) ( x−2 ) ( x−3 )∗x+5(2 x+1 ) ( x−2 )

+(2 x+1 ) ( x−2 ) (x−3 )∗x−1( x−2 ) ( x−3 )

=(2 x+1 ) ( x−2 ) ( x−3 )∗3x+4(2x+1 ) ( x−3 )

Notice the denominators "cleared out", so to speak, leaving only the numerators.

This step is very critical for any rational expression in order to make further

computations easier.

5) ( x−3 ) ( x+5 )+(2 x+1 ) ( x−1 )=( x−2 )(3 x+4) Intermediate result.

At this time, FOIL needs to be performed on both sides of equation in order to

remove grouping symbols, which are the parenthesis.

126

6) x2+5x−3x−15+2 x2−2 x+ x−1=3 x2+4 x−6 x−8 FOIL

7) 3 x2+x−16=3 x2−2 x−8 Like terms collected.

At this time, recognize the presence of a quadratic equation due to 3 x2 term. This

equation must be set equal to zero before further computations can be performed.

Proceed as follows:

8) 3 x2−3 x2+x−16=3 x2−3 x2−2x−8 Subtract3 x2 from each side of

equation.

9) x−16=−2 x−8 Intermediate result, which leaves

only linear equation. The3 x2

term has been cancelled.

At this time, some discussion is required to explain how to solve any linear

equation. The overall strategy is to separate terms containing variables from constants.

In other words, variables go to left side of equals sign, numbers with no variables go to

right side of equals sign.

10) x+2x−16=−2 x+2x−8 Add 2 x to each side of equation.

11) 3 x−16=−8 Intermediate result.

12) 3 x−16+16=−8+16 Add 16to each side of equation.

13) 3 x=8

14) 3x3

=83 Divide each side by3.

127

15) x=83 Final answer


Worksheets contained in packet provided at workshop.


Solve the following quadratic equations.

1) 3 x2=7 x−2

2) 5 x2=9 x

3) ( x+3 ) ( x+2 )=0

4) 3 (5 x+1 ) ( x−7 )=0

5) 7 t ( t−5 )=0

6) x2−8 x=−16

7) 4 t2=25

8) ( x+3 ) (2x−1 )=9

9) 3 x3−30 x=9 x2

10) (2 x−5 ) (3 x2+29 x+56 )=0

Answer key to solving quadratic equations:

128

1) x=13∨x=2

2)x=0∨x=95

3) x=−3∨x=2

4) x=−15

∨x=7

5) t=0∨t=5

6) x=4

7) t=−52

∨t=52

8) x=−4∨x=32

9) x=−2∨x=0∨x=5

10) x=−7∨x=−83

∨x=52


Solve the following rational equations:

1) 2c15

−23=1

3− c

5

129

2) 16c2−4

= c−2c+2

+ c+22−c

3) a3=1a−1

6

4) a−1

2a+5= a−8

6a+15+1

3

5) 1

2c (c+11)+3(c+11)= 1

8c3+27

6) x2+x−12x3−3 x2−2 x+3

= 16 x−9

7) 2 x+3

x2+2 x−3+ x−3

2 x2−3 x+1= 5 x

2 x2+5 x−3

8) x+5

2x2−3 x−2+ x−1x2−5 x+6

= 3 x+42x2−5 x−3

9) x5+ 3x=8

5

10) x3=5

6− 1

2 x

Answer key to solving rational expressions:

1) c=3

2) no solution

130

3) a=−2∨a=32

4) a≠−52

5) c=−14

∨c=2

6) x=12∨x=−2

7) x=43

8) x=83

9) x=3∨x=5

10) x=1∨x=32

References


Boston: Pearson.


July 30, 2010 from

131


12.htm

Bouyssounouse, J. (2010). Sum of two cubes worksheet. www.edonyourown.com.

Retrieved August 1, 2010 from http://www.edonyourown.com/

Duval, R. Useful tips to solving algebra word problems. (2008). www.articlesbase.com.

Retrieved August 1, 2010 from http://www.articlesbase.com/

Factoring polynomials having common factors. (n.d.). www.lessonplanet.com. Retrieved

October 31, 2010 from http://blue.utb.edu/jtelese/example%20lessons.htm






Russell, D. (n.d.).www.about.com.Retrieved July 15, 2010 from

http://math.about.com/library/bldivide.htm



Company.



132

End of Section

133

Appendix B: Student Packet

FORMULAS FOR FACTORING WORKSHOPS

LET A= FIRST TERM LET B= LAST TERM

134

SUM OF PERFECT SQUARES: A2+B2=¿CANNOT FACTOR, PRIME (IF

YOU ARE IN ELEMENTARY OR INTERMEDIATE LEVEL ALGEBRA

ONLY)

DIFFERENCE OF PERFECT SQUARES: A2−B2=(A−B)(A+B)

SUM OF CUBES: A3+B3=(A+B)(A2−AB+B2)

DIFFERENCE OF CUBES: A3−B3=(A−B)(A2+AB+B2)

MONIC-TRINOMIAL: x2+bx+c

NON-MONIC TRINOMIAL: a x2+bx+c a≠1 ,0

PERFECT SQUARE TRINOMIAL: ( A+B )2=A2+2 AB+B2

( A−B )2=A2−2 AB+B2

Bittinger, Ellenbogen, & Johnson (2006)

135

FORMULAS FOR FACTORING WORKSHOPS

SOME TEACHERS PREFER USING THIS FORMAT

LET F= FIRST TERM LET L= LAST TERM

SUM OF PERFECT SQUARES: F2+L2=¿ CANNOT FACTOR, PRIME (IF

YOU ARE IN ELEMENTARY OR INTERMEDIATE LEVEL ALGEBRA

ONLY)

DIFFERENCE OF PERFECT SQUARES: F2−L2=(F−L)(F+ L)

SUM OF CUBES: F3+L3=(F+L)(F2−FL+L2)

DIFFERENCE OF CUBES: F3−L3=(F−L)(F2+FL+L2)

MONIC-TRINOMIAL: x2+bx+c

NON-MONIC TRINOMIAL: a x2+bx+c a≠1 ,0

PERFECT SQUARE TRINOMIAL: (F+ L )2=F2+2 FL+L2

(F−L )2=F2−2FL+L2

Permission from Hall, E. (2010)

136

TO FACTOR A POLYNOMIAL

1) ALWAYS LOOK FOR A COMMON FACTOR, TERM,

EXPRESSION OR VARIABLE FIRST. IF THERE IS ONE,

FACTOR OUT THE GREATEST COMMON FACTOR

(GCF). BE SURE TO INCLUDE IT IN THE FINAL ANSWER.

2) LOOK AT THE NUMBER OF TERMS:

TWO TERMS: DIFFERENCE OF SQUARES, SUM OF

SQUARES, SUM OF CUBES, DIFFERENCE OF

CUBES, PERFECT SQUARE TRINOMIAL.

THREE TERMS: MONIC TRINOMIAL, NON-MONIC

TRINOMIAL, OR PERFECT SQUARE TRINOMIAL.

FOUR TERMS: TRY FACTORING BY GROUPING.

3) ALWAYS FACTOR COMPLETELY!!!

4) TO CHECK YOUR WORK, DO FOIL!!!!!

Bittinger, Ellenbogen, & Johnson (2006)

137

PERFECT SQUARES PERFECT CUBES12=1∗1=1 13=1∗1∗1=122=2∗2=4 23=2∗2∗2=832=3∗3=9 33=3∗3∗3=27

42=4∗4=16 43=4∗4∗4=6452=5∗5=25 53=5∗5∗5=12562=6∗6=36 63=6∗6∗6=21672=7∗7=49 73=7∗7∗7=34382=8∗8=64 83=8∗8∗8=51292=9∗9=81 93=9∗9∗9=729

102=10∗10=100 103=10∗10∗10=1000112=11∗11=121 113=11∗11∗11=1331122=12∗12=144 123=12∗12∗12=1728132=13∗13=169 133=13∗13∗13=2197142=14∗14=196 143=14∗14∗14=2744152=15∗15=225 153=15∗15∗15=3375162=16∗16=256 163=16∗16∗16=4096172=17∗17=289 173=17∗17∗17=4913182=18∗18=324 183=18∗18∗18=5832192=19∗19=361 193=19∗19∗19=6859202=20∗20=400 203=20∗20∗20=8000212=21∗21=441 213=21∗21∗21=9261222=22∗22=484 223=22∗22∗22=10648232=23∗23=529 233=23∗23∗23=12167242=24∗24=576 243=24∗24∗24=13824252=25∗25=625 253=25∗25∗25=15625262=26∗26=676 263=26∗26∗26=17576272=27∗27=729 273=27∗27∗27=19683282=28∗28=784 283=28∗28∗28=21952292=29∗29=841 293=29∗29∗29=24389302=30∗30=900 303=30∗30∗30=27000312=31∗31=961 313=31∗31∗31=29791

322=32∗32=1024 323=32∗32∗32=32768332=33∗33=1089 333=33∗33∗33=35937342=34∗34=1156 343=34∗34∗34=39304352=35∗35=1225 353=35∗35∗35=42875362=36∗36=1296 363=36∗36∗36=46656372=37∗37=1369 373=37∗37∗37=50653382=38∗38=1444 383=38∗38∗38=54872392=39∗39=1521 393=39∗39∗39=59319402=40∗40=1600 403=40∗40∗40=64000412=41∗41=1681 413=41∗41∗41=68921

138

422=42∗42=1764 423=42∗42∗42=74088432=43∗43=1849 433=43∗43∗43=79507442=44∗44=1936 443=44∗44∗44=85184

Author (2010)

LIST OF PRIME NUMBERS

2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 631 641 643 647 653 659 661 673 677 683 691 701 709 719 727 733 739 743 751 757 761 769 773 787 797 809 811 821 823 827 829 839 853 857 859 863 877 881 883 887 907 911 919 929 937 941 947 953 967 971 977 983 991 997 1009 1013 1019 1021 1031 1033 1039 1049 1051 1061 1063 1069 1087 1091 1093 1097 1103 1109 1117 1123 1129 1151 1153 1163 1171 1181 1187 1193 1201 1213 1217 1223 1229 1231 1237 1249 1259 1277 1279 1283 1289 1291 1297 1301 1303 1307 1319 1321 1327 1361 1367 1373 1381 1399 1409 1423 1427 1429 1433 1439 1447 1451 1453 1459 1471 1481 1483 1487 1489 1493 1499 1511 1523 1531 1543 1549 1553 1559 1567 1571 1579 1583 1597 1601 1607 1609 1613 1619 1621 1627 1637 1657 1663 1667 1669 1693 1697 1699 1709 1721 1723 1733 1741 1747 1753 1759 1777 1783 1787 1789 1801 1811 1823 1831 1847 1861 1867 1871 1873 1877 1879 1889 1901 1907 1913 1931 1933 1949 1951 1973 1979 1987 1993 1997 1999 2003 2011 2017 2027 2029 2039 2053 2063 2069 2081 2083 2087 2089 2099 2111 2113 2129 2131 2137 2141 2143 2153 2161 2179 2203 2207 2213 2221 2237 2239 2243 2251 2267 2269 2273 2281 2287 2293 2297 2309 2311 2333 2339 2341 2347 2351 2357 2371 2377 2381 2383 2389 2393 2399 2411 2417 2423 2437 2441 2447 2459 2467 2473 2477 2503 2521 2531 2539 2543 2549 2551 2557 2579 2591 2593 2609 2617 2621 2633 2647 2657 2659 2663 2671 2677 2683 2687 2689 2693 2699 2707 2711 2713 2719 2729 2731 2741 2749 2753 2767 2777 2789 2791 2797 2801 2803 2819 2833 2837 2843 2851 2857 2861 2879 2887 2897 2903 2909 2917 2927 2939 2953 2957 2963 2969 2971 2999 3001 3011 3019 3023 3037 3041 3049 3061 3067 3079 3083 3089 3109 3119 3121 3137 3163 3167 3169 3181

139

3187 3191 3203 3209 3217 3221 3229 3251 3253 3257 3259 3271 3299 3301 3307 3313 3319 3323 3329 3331 3343 3347 3359 3361 3371 3373 3389 3391 3407 3413 3433 3449 3457 3461 3463 3467 3469 3491 3499 3511 3517 3527 3529 3533 3539 3541 3547 3557 3559 3571 3581 3583 3593 3607 3613 3617 3623 3631 3637 3643 3659 3671 3673 3677 3691 3697 3701 3709 3719 3727 3733 3739 3761 3767 3769 3779 3793 3797 3803 3821 3823 3833 3847 3851 3853 3863 3877 3881 3889 3907 3911 3917 3919 3923 3929 3931 3943 3947 3967 3989 4001 4003 4007 4013 4019 4021 4027 4049 4051 4057 4073 4079 4091 4093 4099 4111 4127 4129 4133 4139 4153 4157 4159 4177 4201 4211 4217 4219 4229 4231 4241 4243 4253 4259 4261 4271 4273 4283 4289 4297 4327 4337 4339 4349 4357 4363 4373 4391 4397 4409 4421 4423 4441 4447 4451 4457 4463 4481 4483 4493 4507 4513 4517 4519 4523 4547 4549 4561 4567 4583 4591 4597 4603 4621 4637 4639 4643 4649 4651 4657 4663 4673 4679 4691 4703 4721 4723 4729 4733 4751 4759 4783 4787 4789 4793 4799 4801 4813 4817 4831 4861 4871 4877 4889 4903 4909 4919 4931 4933 4937 4943 4951 4957 4967 4969 4973 4987 4993 4999 5003 5009 5011 5021 5023 5039 5051 5059 5077 5081 5087 5099 5101 5107 5113 5119 5147 5153 5167 5171 5179 5189 5197 5209 5227 5231 5233 5237 5261 5273 5279 5281 5297 5303 5309 5323 5333 5347 5351 5381 5387 5393 5399 5407 5413 5417 5419 5431 5437 5441 5443 5449 5471 5477 5479 5483 5501 5503 5507 5519 5521 5527 5531 5557 5563 5569 5573 5581 5591 5623 5639 5641 5647 5651 5653 5657 5659 5669 5683 5689 5693 5701 5711 5717 5737 5741 5743 5749 5779 5783 5791 5801 5807 5813 5821 5827 5839 5843 5849 5851 5857 5861 5867 5869 5879 5881 5897 5903 5923 5927 5939 5953 5981 5987 6007 6011 6029 6037 6043 6047 6053 6067 6073 6079 6089 6091 6101 6113 6121 6131 6133 6143 6151 6163 6173 6197 6199 6203 6211 6217 6221 6229 6247 6257 6263 6269 6271 6277 6287 6299 6301 6311 6317 6323 6329 6337 6343 6353 6359 6361 6367 6373 6379 6389 6397 6421 6427 6449 6451 6469 6473 6481 6491 6521 6529 6547 6551 6553 6563 6569 6571 6577 6581 6599 6607 6619 6637 6653 6659 6661 6673 6679 6689 6691 6701 6703 6709 6719 6733 6737 6761 6763 6779 6781 6791 6793 6803 6823 6827 6829 6833 6841 6857 6863 6869 6871 6883 6899 6907 6911 6917 6947 6949 6959 6961 6967 6971 6977 6983 6991 6997 7001 7013 7019 7027 7039 7043 7057 7069 7079 7103 7109 7121 7127 7129 7151 7159 7177 7187 7193 7207 7211 7213 7219 7229 7237 7243 7247 7253 7283 7297 7307 7309 7321 7331 7333 7349 7351 7369 7393 7411 7417 7433 7451 7457 7459 7477 7481 7487 7489 7499 7507 7517 7523 7529 7537 7541 7547 7549 7559 7561 7573 7577 7583 7589 7591 7603 7607 7621 7639 7643 7649 7669 7673 7681 7687 7691 7699 7703 7717 7723 7727 7741 7753 7757 7759 7789 7793 7817 7823 7829 7841 7853 7867 7873 7877 7879 7883 7901 7907 7919

140

Permission from Caldwell, C. (2009)

141

Rules of Divisibility

1 divides every whole number

2 divides a whole number if it is even

3 divides a whole number if three divides the sum of the digits

4 divides a whole number if four divides the last two digits

5 divides a whole number if it ends in five or zero

6 divides a whole number if the rules for two and three above both work

7 divides a whole number if removing the digit from the ones position, doubling it, and subtracting it from the rest of the number, yields a number divisible by seven

8 divides a whole number if eight divides the last three digits

9 divides a whole number if nine divides the sum of the digits

10 divides a whole number if it ends in zero

Permission from Young, E. (2009)

142

Permission from Williams, J. (2010)

144

Name: Date:

Divisibility RulesNumber Divisible

By

My Hypothesis The Actual Rule

2

3

4

5

6

7

8

9

10

Permission from Russell, D. (n.d.)

145

Name___________________

Divisibility Practice

Determine if each number is divisible by 2, 3, 4, 5, 6, 7, 8, 9 or 10.

Make a list under each number.

1.) 210 2.) 72 3.) 145

4.) 333 5.) 1400 6.) 62

7.) 1008 8.) 2310 9.) 144

10.) 231

Permission from Russell, D. (n.d.)

146

Greatest Common Factor

1) 9a3 x−6 a2 x4+12a5 x4

2) 6a (2 x−5 )−7 (2x−5 )

3) 3 x2 (7 c−4 )−2x (7c−4 )+5 (7c−4 )

4) −10c5 x3−15c3 x+5 c2 x

5) 5 (3a−4 )−15c (3 a−4 )+10u (3a−4 )

147

Factoring By Grouping Method

1) 15ac−6 c+20a−8

2) 6mx−4mz−3x+2 z

3) 15aw2+5 cw 2−6a−2c

4) y6− y5− y+1

5) 10a2c+5a2x−6 c−3 x

148

Difference of Squares

Formula:F2−L2=(F−L)(F+ L)

1) x2−256

2) x2−625

3) x2−1024

4) 100− x2

5) x4−10000

149

Sum of Squares

Formula:F2+L2=prime

1) y2+529

2) b2+1225

3) w2+256

4) x2+121

5) a2+2500

150

Sum of Cubes

Formula: F3+L3=(F+L)(F2−FL+L2)

1) 27 x3+64

2) 8c3+x3

3) x6+8

4) 1000+x9

5) h6+q6

151

Difference of Cubes

Formula:F3−L3=(F−L)(F2+FL+L2)

1) t 3−1000

2) 27 x3−64 z3

3) h3− j3

4) 64−r9

5) 512 c3−z6

152

Monic Trinomials

x2+bx+c

1) x2−5 x−14

2) x2+9 x+14

3) z2−44 z+84

4) z2+40 z−225

5) k 2−49k−392

153

Non-Monic Trinomials

a x2+bx+c a≠1 ,0

1) 2 x2+15 x+28

2) 8a2−26a+15

3) 33 z2−4 z−37

4) 81h2+225h+100

5) 48 d2−19d−67

154

Solve by Factoring

1) 2 x3−7 x2+6 x=0

2) 4m3+5m2=6m

3) x2+ x−2=4

4) (g−1 ) (g+2 )=10

5) 4 x

2x2+5 x−3+ 3x2+3x

= 12x2−x

155

Putting It All Together

1) 10a (2 x+7 )+15(2 x+7) 2) 49 x2−25 z2

3) 81 z4+100 c4 4) z2−71 z−810

5) 25c2+34c−32 6) x6−4096

7) z3+64 8) 6h3+12h=17h2

9) 1

2c (c+11)+3 (c+11 )= 1

8c3+2710)

518c

− 49c

+ 12c

11) 3b+3

10 x+15∗2 x+3

5b+512) 2c2−4c

6 c−6÷ c3−4 cc2−2c−8

13) 3x

x2+7 x+9− 3 x+5x2+7 x+9

14) ( 5x+2

− 3x−2 )

xx−1

+ xx+1

156

References


Boston: Pearson.

Caldwell, C. (2009). The primes page. Retrieved August 1, 2010 from University of

Tennessee at Martin: http://primes.utm.edu/lists/small/1000.txt






Company.

Young, E. (2009). Math 5325: Structure of Number Concepts. Divisibility Rules.

Retrieved June 15, 2009 from Texas A&M University, Corpus Christi, TX:

http://sci.tamucc.edu/~eyoung/1350/divisibility.html

Williams, J. (n.d.) Factoring polynomial flow chart. Retrieved August 15, 2010 from

http://www.schools.manatee.k12.fl.us/072JWILLIAMS/072jwilliams1/factoring_pol

ynomials_flowchart.html

157

Appendix C: Student Packet Answer Key

158

Greatest Common Factor

1) 3 xa2 (3a−2 x3+4 a3 x3 ) 2) (2 x−5 ) (6a−7 )

3) (7c−4)(3 x2−2 x+5) 4) −5c2 x (2c3 x2+3c−1)

5) 5(3 a−4)(1−3c+2u)

Factoring by Grouping Method

1) (5a−2 ) (3 c+4 ) 2) (3 x−2 z)(2m−1)

3) (3a+c)(5w2−2) 4) ( y−1)( y5−1)

5) (2c+x)(5c2−3)

Difference of Perfect Squares

1) (x−16)(x+16) 2) (x−25)(x+25)

3) (x−32)(x+32) 4) (10−x)(10+x )

5) (x−10)(x+10)(x2+100)

159

Sum of Squares

1) y=±23 i 2) b=±35i

3) w=±16 i 4) x=±11 i

5) a=±50i

Sum of Cubes

1) (3 x+4)(9x2−12 x+16) 2) (2c+x)(4c2−2cx+x2)

3) (x2+2)(x4−2 x2+4) 4)(10+x3)¿)

5) (h2+q2)(h4−h2q2+q4)

Difference of Cubes

1) (t−10)(t 2+10 t+100) 2) (3 x−4 z)(9x2+12xz+16 z2)

3) (h− j)(h2+hj+ j2) 4) (4−r3)(16+4 r3+r6)

5) (8c−z2)(64c2+8 c z2+z4)

160

Monic Trinomials

1) ( x−7 )(x+2) 2) ( x+7 )(x+2)

3) ( z−42 )(z−2) 4) ( z+45 )(z−5)

5) (k−56 )(k+7)


1) (2 x+7 ) ( x+4 ) 2) (4 a−3 ) (2a−5 )

3) (33 z−37 )( z+1) 4) (9h+5 )(9h+20)

5) (48 d−67 )(d+1)

Solve by Factoring

1) x=0∨x=32∨x=2 2) m=−2∨m=0∨m=3

4

3) x=−3∨x=2 4) g=−4∨g=3

5) x=−2∨x=34

161

Putting It All Together

1) 5(2 x+7)(2a+3) 2) (7 x−5 z)(7 x+5 z)

3) cannot factor , prime 4) (z+10)(z−81)

5) (25 x−16)(x+2) 6) (x−4)(x+4)(x 4+16 x2+256)

7) (z+4 )(z2−4 z+16) 8) h=0∨h=34∨h=3

2

9) c=−14

∨c=2 10) 1

3c

11) 3

25 12) c−4

3 (c−1 )

13) −5

x2+7 x+914)

( x−8 ) ( x−1 ) (x+1 )x2 ( x−2 ) ( x+2 )

162

References

Bittinger, E. (2006). Elementary and Intermediate Algebra,Concepts and Applications.

Boston: Pearson.

Hall, E. (2010). College Algebra. Corpus Christi: unknown.

Hall, E. (2010). Elementary Algebra. Corpus Christi: unknown.

Hall, E. (2010). Intermediate Algebra. Corpus Christi: unknown.



Company.

163

Appendix D: Hyperlinks for Videos

Title: Introduction to Factoring VideosLink: http://mars.delmar.edu/mediasite/viewer/?peid=285b809d-1e0f-414d-906e-578030d322fb

Title: Divisibility RulesLink: http://mars.delmar.edu/mediasite/viewer/?peid=f0d89401-9207-4361-aa9c-c9b10d6a95bd

Title: Greatest Common Factor (GCF)Link: http://mars.delmar.edu/mediasite/viewer/?peid=2d94d44b-9ebb-4256-b4ba-859e6c6bf9fc

Title: Factoring by GroupingLink: http://mars.delmar.edu/mediasite/viewer/?peid=61dff3a1-5f2d-4836-b198-4513eece98f0

Title: Difference of Perfect SquaresLink: http://mars.delmar.edu/mediasite/viewer/?peid=0e9cc95b-89a6-4d7b-8fd3-13ebfd354ef8

Title: Sum of SquaresLink: http://mars.delmar.edu/mediasite/viewer/?peid=1f158fa2-f8cb-4220-b648-68afc30cb3d6

Title: Sum of CubesLink: http://mars.delmar.edu/mediasite/viewer/?peid=7e5fabcf-c986-4a87-8bcb-e8945b9866d2

Title: Difference of CubesLink: http://mars.delmar.edu/mediasite/viewer/?peid=a5392dba-506c-4bde-829c-9a56510c6e4e

Title: Monic TrinomialsLink: http://mars.delmar.edu/mediasite/viewer/?peid=d2d59472-a661-476b-ad68-f3ff03a1f741

Title: Non-Monic TrinomialsLink: http://mars.delmar.edu/mediasite/viewer/?peid=4e53504a-8de9-450d-8a3e-fe08c9cf5564

Title: Solve by FactoringLink: http://mars.delmar.edu/mediasite/viewer/?peid=7518c534-a431-474e-a3a7-4860c45a508d

Title: Sum of Squares College Algebra Students OnlyLink: http://mars.delmar.edu/mediasite/viewer/?peid=a6d49adc-f6eb-4dd7-a040-830a87f7a5e4

164

Appendix E: Additional Problems and Solutions Available for Students

165

Greatest Common Factor Problems

Factor the following:

1) 9a3 x−6 a2 x4+12a5 x4

2) 6a (2 x−5 )−7 (2x−5 )

3) 18a5 x+9a3 x2−12a2 x4

4) 12c4 y6+6c2 y3+24 c5 y4

5) −5m2 x5−8m x4+3 x2

6) 7a (3c+2 )−5 (3c+2 )

7) 3 x2 (7 c−4 )−2x (7c−4 )+5 (7c−4 )

8) 6 x (a+1 )−9 (a+1 )

9) 12c (3h−5 )+6 (3h−5 )

10) 15a (3 y+1 )−5 (3 y+1 )+35c (3 y+1 )

11) −10c5 x3−15c3 x+5 c2 x

12) 20a (3 x+1 )−10c (3 x+1 )

13) 36c2m3−48c3m+24 c2m2

14) −40k5 y−16k4 y2+24k 3 y3−8k2 y 4

15) 8 x (3 v+2 )−7 (3v+2 )

16) 7 p2 (m−5 z )+t 3 (m−5 z )

17) 8a3 (5m−2 )−12a2 (5m−2 )+20a (5m−2 )

18) 3a+6

19) 5ab−15b2

20) 6 x2 y−9 x3+3 x

21) −3 c3 x+18c4 x2−6cx

166

22) a5b2c3−a2b2c2+ab4 c4

23) −25a4 x+5a3 x2+15a2 x3

24) 14a x2−7 a3+21ax3

25) 3a ( x+2 y )+5 ( x+2 y )

26) 8 y2 (4 x+5 )+3 y (4 x+5 )−2 ( 4 x+5 )

27) 13d3 (3 c−z )−2d2 (3c−z )+5 (3c−z )

28) 4 a2 (2x−3 )−6 (2 x−3 )

29) (5c+1 )−2 x (5c+1 )

30) 3 x3+6x2−12 x

31) 8a−12

32) 24 x5+30 x2

33) 12 x5−15x4+27x3

34) 5 x−15

35) 2 t2+t

36) −6a2−9a

37) 6 x8+12x6−24 x4+30 x2

38) x2 ( x−7 )−3 ( x−7 )

39) 5 x5−5 x4+x3−x2+3 x−3

40) 5 (3a−4 )−15c (3 a−4 )+10u (3a−4 )

41) 9 x (a+b )+12 y (a+b )−3 (a+b )

42) 14 (3a+x )−4 a (3 a+x )

43) x2 ( x+1 )+2 (x+1 )

44) a2b2+15b2

167

45) 8 x5+26 x7

46) 5a2 x−15 x

47) y ( y−2 )+7 ( y−2 )

48) b (b−7 )+(b−7 )

49) 50u2+88u x2

References

Bittinger, E. (2006). Elementary and Intermediate Algeba.Concepts and Applications.

Boston: Pearson.






Company.

Solutions to previous problems on the following pages.

168

Greatest Common Factor Solutions

1) 3 xa2 (3a−2 x3+4 a3 x3 )

2) (2 x−5 ) (6a−7 )

3) 3a2 x¿

4) 6c2 y3 (2c2 y3+1+4c3 y )

5) – x2 (5m2 x3+8mx2−3 )

6) (3c+2)(7a−5)

7) (7c−4)(3 x2−2 x+5)

8) (a+1)(6 x−9)

3(a+1)(2x−3)

9) (3h−5)(12c+6)

6(3h−5)(2c+1)

10) (3 y+1)(15a−5+35 c)

5(3 y+1)(5a−1+7c)

11) −5c2x (2c3 x2+3c−1)

12) (3 x+1)(20a−10c )

10(3 x+1)(2a−c )

13) 12c2m(3m2+4 c+2m)

14) −8 k 2 y (5k3+2k2 y−3k y2+ y3)

15) (3v+2)(8 x−7)

16) (m−5 z )(7 p2+ t3)

17) (5m−2)(8a3−12a2+20a)

169

4 a(5m−2)(2a2−3a+5)

18) 3(a+2)

19) 5b(a−3b)

20) 3 x (2xy−3 x2+1)

21) 18 c4 x2−3c3x−6 cx

3cx (6c3 x−c2−2)

22) ab2c2(a4c−a+b2 c2)

23) −5a2 x (5a2−ax−3 x2)

24) 7a (2x2−a2+3x3)

25) (x+2 y)(3a+5)

26) (4 x+5)(8 y2+3 y−2)

27) (3c−z )(13d3−2d2+5)

28) (2 x−3)(4 a2−6)

2(2 x−3)(2a2−3)

29) (5c+1)(1−2 x)

30) 3 x (x2+2 x−4)

31) 4 (2a−3)

32) 6 x2(4 x3+5)

33) 3 x3(4 x2−5 x+9)

34) 5(x−3)

35) 2 t( t+1)

36) −3a(2a+3)

170

37) 6 x2(x6+2x4−4 x2+5)

38) (x−7)(x2−3)

39) (5 x5−5 x4 )+(x3−x2 )+(3x−3)

5 x4 ( x−1 )+ x2 ( x−1 )+3(x−1)

(x−1)(5 x4+x2+3)

40) (3a−4 )(5−15c+10u)

5(3 a−4)(1−3c+2u)

41) (a+b)(9 x+12 y−3)

3(a+b)(3 x+4 y−1)

42) (3a+x )(14−4 a)

2(3a+ x)(7−2a)

43) (x+1)(x2+2)

44) b2(a2+15)

45) 2 x5(4+13 x2)

46) 5 x (a2−3)

47) ( y−2)( y+7)

48) (b−7)(b+1)

49) 2u(25u+44 x2)

References


Boston: Pearson.



171




Company.

172

Additional Greatest Common Factor Problems with Detailed Solutions

Recall that to sum means to add. The expressions to be added are called terms.

In the expression,6ab−78 x+15 x3 there are three terms. The first term is6ab, the second

term is−78 x, and the third term is15 x3.

To find a product means to multiply. The numbers or variables being multiplied

are called factors.

The expression5ab (2x+7)(3 x2−8 x+1) is considered to be in factored form,

where (3 x2−8 x+1 ) is a trinomial factor, (2 x+7 )is a binomial factor, and 5ab is called a

monomial factor.

(Hall, 2010)

Ask the following questions first when confronted with any type of factoring

problem or equation:

1) Is there a greatest common factor (GCF) present? If there is one, factor out

the GCF.

2) How many terms are present? Are there two, three or four terms present in the

problem or equation?

173

Factoring an algebraic

expression is writing terms as a

product

3) Is a difference of perfect squares or sum of perfect squares present? Is a

difference of cubes or sum of cubes present?

4) Are three terms present? Consider factoring a monic trinomial or non-monic

trinomial.

5) Are four terms present? Consider factoring by the grouping method.

Greatest Common Factor Techniques

This section deals with factoring the greatest common factor from an algebraic

expression or equation. Simply stated, “What number, variable, or complete expression

do all terms have in common?”

Example:9a3x−6 a2 x4+12a5 x4

Decompose or break down each term completely on paper until proficiency is

attained. Proceed as follows:


1) 9a3 x−6 a2 x4+12a5 x4 Recopy problem.

2) 3a2 x∗3a−3a2 x∗2 x3+3a2 x∗4a3 x3 Decompose, or factor each term.

The following expression was “broken down” or decomposed in order to see the

common factor, which is3a2 x. Proceed as follows:

3) 3a2 x (3a−2 x3+4 a3 x3) Greatest common factor is written down

first, followed by remaining terms

enclosed within parenthesis.

4) 3a2 x (3a−2 x3+4 a3 x3) Final answer.

174

Example:12c4 y6+6c2 y3+24 c5 y4

Ask the following questions first when confronted with any type of factoring:


the GCF.






trinomial.





do all of the terms have in common?”




1) 12c4 y6+6c2 y3+24 c5 y4 Recopy problem.

2) 6c2 y3∗2c2 y3+6c2 y3∗1+6 c2 y3∗4 c3 y Decompose, or factor each term.

175

Ask “What is the greatest common factor that is present in all terms?” The

common factor, number or variable that is present in all terms is6c2 y3. Proceed as

follows:

3) 6c2 y3(2c2 y3+1+4 c3 y) Greatest common factor is

written down first followed by

remaining terms enclosed in

parenthesis.

4) 6c2 y3(2c2 y3+1+4 c3 y) Final answer.

Be aware of the fact that a factor of 1 is always present and is assumed to be

attached to any term in math. It is implied in mathematics, but not easily seen by

students.

176

Example: −10 t5+15 t 4+9 t 3

Ask the following questions first when confronted with any type of

factoring:


the GCF.






trinomial.









1) −10 t5+15 t 4+9 t 3 Recopy problem.

2) −5∗2∗t 3∗t2+5∗3∗t 3∗t+3∗3∗t 3 Decompose, or factor each term.

3) −10∗t3∗t2+15∗t 3∗t+9∗t3 t 3is common factor.

177

4) −t 3(10 t2−15 t−9) Factor out−t 3.

It is customary to factor out a “negative” term when the first leading coefficient is

negative. When this is performed, the signs of all terms will change.

5) −t 3(10 t2−15 t−9) Final answer.

178

Example:7a (3c+2 )−5 (3c+2)



the GCF.






trinomial.






For this type of problem,(3c+2) is considered a factor and cannot be separated

due to presence of encompassing parenthesis. Proceed as follows:


1) 7a (3c+2 )−5 (3c+2) Recopy problem.

2) 7a (3c+2 )−5 (3c+2) (3c+2 )is common factor.

3) (3c+2 )(7a−5) Common factor written first,

followed by monomial factors.

4) (3c+2)(7a−5) Final answer.

179

Example:15a (3 y+1 )−5 (3 y+1 )+35c (3 y+1 )

Ask the following questions first when confronted with any type of

factoring:


the GCF.






trinomial.






For this type problem, there are three expressions the same. The expression

(3 y+1) is considered the common factor and is construed to be a single unit which

cannot be separated due to the parenthesis. Proceed as follows:


1) 15a (3 y+1 )−5 (3 y+1 )+35c (3 y+1 ) Recopy problem.

2) 15a (3 y+1 )−5 (3 y+1 )+35c (3 y+1 )

180

3)(3 y+1)(15a−5+35 c) Common factor of (3 y+1)

written first, followed by

remaining terms.

At this time, the three term expression still contains a common factor of5.

Decompose, or factor only the last expression, that is,(15 a−5+35c ).

4) (3 y+1 )(5∗3a−5∗1+5∗7c) Decompose last expression only.

5)(3 y+1 )(5∗3a−5∗1+5∗7c) Notice the common factor of 5 in

last expression.

6)(3 y+1 )5 (3a−1+7 c) Factor 5 from last expression.

One more step is needed to finish process and that is to bring 5 to front of

expression.

Commutative Law

ab=ba (Bittinger, 2006)

7) 5 (3 y+1 )(3a−1+7 c) Customary to present coefficient

or number in front of answer.

8) 5 (3 y+1 )(3a−1+7 c) Final answer.

References


Boston: Pearson.





181


Company.

End of Section

182

Factoring by Grouping Method

Factor the following:

1) 15ac−6 c+20a−8

2) 15a2+10ay−12ax−8xy

3) 6mx−4mz−3x+2 z

4) 2c−4 y+3cy−6 y2

5) 4 a−12−5ab+15b

6) 12a2x−8a−3ax+2

7) 2dq−10d+q−5

8) 6 vw−9v+8w−12

9) 15aw2+5 cw 2−6a−2c

10) 8 xy−28 x+2 y−7

11) y6− y5− y+1

12) 6ax+2ay+9 x+3 y

13) 8ac−4c−6a+3

14) 3 x2−6 x+5ax−10a

15) 12ac2+8c3+6ax+4cx

16) acx−cxy−a+ y

17) 2a2 x+15+5 x+6a2

18) 3 xy−6 y2+x−2 y

19) 8ac−4c−6a+3

20) 10a2 c+5a2 x−6c−3 x

21) 3ax−6 ay+ x−2 y

183

22) 5bc2−5bc−c+1

23) 5 x+6a−2ax−15

24) 5a2−3a−1+15a3

References


Boston: Pearson.






Company.


184

Factoring by Grouping Method Solutions

1) (5a−2 ) (3 c+4 )

2) (3a+2 y )(5a−4 x )

3) (3 x−2 z)(2m−1)

4) (c−2 y )(2+3 y)

5) (a−3)(4−5b)

6) (4a−1)(3 ax−2)

7) (q−5)(2d+1)

8) (2w−3)(3 v+4)

9) (3a+c)(5w2−2)

10) (4 x+1)(2 y−7)

11) ( y−1)( y5−1)

12) (3 x+ y)(2a+3)

13) (2a−1)(4 c−3)

14) (x−2)(3 x+5a)

15) (3a+2)(4c2+2x )

16) (a− y )(cx−1)

17) (2a2+5 ) ( x+3 )

18) (x−2 y )(3 y+1)

19) (2a−1)(4 c−3)

20) (2c+x)(5c2−3)

21) (x−2 y )(3a+1)

22) (c−1)(5bc−1)

185

23) (5−2a)( x−3)

24) (3a+1)(5a2−1)

References


Boston: Pearson.






Company.

186

Additional Factoring by Grouping Method Problems with Detailed Solutions



the GCF.






trinomial.


This section deals with factoring by grouping, which is usually limited to four

terms. The overall strategy for this technique is to group in pairs while simultaneously

factoring out the greatest common monomial factor.

187

It is important to note that a plus

or minus sign must precede

(come before) the second

grouping.

Example: 2 x3+8 x2+x+4


1) (2 x3+8x2)+(x+4) Recopy problem, group first two terms,

and then group last two terms.

2) (2 x2∗x+2x2∗4 )+(x+4 ) Decompose, or factor terms.

Factor out common factor from first two terms, which is2 x2; the last two terms

have no common factor except for an assumed factor of1.

3) 2 x2 ( x+4 )+(x+4 )

4) 2 x2 ( x+4 )+1(x+4) Assumed factor of1.

5) 2 x2 ( x+4 )+1(x+4)

6) (x+4 )(2x2+1) Common factor of (x+4 ) is inserted

first, then remaining two terms are

inserted last.

7) (x+4 )(2x2+1) Final answer.

188

Example:8 x4−28x3+6 x−21



the GCF.






trinomial.




factoring out the greatest common monomial factor (Hall, 2010).

(Hall, 2010)

189




grouping.

Four terms are present, asserting factoring by grouping method. Proceed as

follows:


1) 8 x4−28x3+6 x−21 Recopy problem.

2) (8 x4−28x3)+(6 x−21) Group first two terms, then last

two terms.

3) (4 x3∗2 x−4 x3∗7)+(3∗2x−3∗7) Decompose each term for better

visualization of common factors.

4) 4 x3 (2 x−7 )+3(2 x−7) Factor GCF’s.

5) 4 x3 (2 x−7 )+3(2 x−7)

6) (2 x−7)(4 x3+3) Common factor is placed first,

and then remaining two terms are

placed second.

7) (2 x−7)(4 x3+3) Final answer.

190

Example: 20 g3−4 g2−25 g+5



the GCF.






trinomial.





(Hall, 2010)


follows:


1) 20 g3−4 g2−25 g+5 Recopy problem.

191




grouping.

2) (20 g3−4 g2 )+(−25 g+5) Group first two terms, then group

last two terms, emphasizing the

placement of a plus sign between

second term and third term.

3) (20 g3−4 g2 )−1(25g−5) Factor out a negative 1 from only

last two terms to ease further

operations and manipulations.

Notice how signs change!

4) ( 4 g2∗5 g−4 g2∗1 )−(5∗5 g−5∗1) Decompose, or factor each term.

5) 4 g2 (5 g−1 )−5(5 g−1) Factor GCF’s.

6) 4 g2 (5 g−1 )−5(5 g−1)

7) (5 g−1)(4 g2−5) Common factor is placed first,

and then remaining two terms

placed second.

8) (5 g−1)(4 g2−5) Final answer.

192

Example: 2 x3−8 x2−9 x+36



the GCF.






trinomial.





(Hall, 2010)


follows:


1) 2 x3−8 x2−9 x+36 Recopy problem.

193




grouping.

2) (2 x3−8 x2)+(−9x+36) Group first two terms, then group

last two terms, emphasizing the

placement of a plus sign between

second term and third term.

3) (2 x3−8 x2)−1(9 x−36) Factor out a negative 1

from only last two terms to ease

further operations and

manipulations. Notice how

signs change!

4) (2 x2∗x−2 x2∗4 )−(9∗x−9∗4 ) Decompose, or factor each term.

5) 2 x2 ( x−4 )−9(x−4) Factor out GCF’s.

7) 2 x2 ( x−4 )−9(x−4)

6) (x−4)(2 x2−9) Common factor placed first, and

then remaining terms placed

second.

7) (x−4)(2 x2−9) Final answer.

194

References


Boston: Pearson.






Company.

End of Section

195

Difference of Perfect Squares

1) x2−1 2) x2−4 3) x2−9

4) x2−16 5) x2−25 6) x2−36

7) x2−49 8) x2−64 9) x2−81

10) x2−100 11)x2−121 12) x2−144

13) x2−169 14) x2−196 15) x2−225

16) x2−256 17) x2−289 18)x2−324

19) x2−361 20) x2−400 21)x2−441

22) x2−484 23) x2−529 24¿ x2−576

25) x2−625 26) x2−676 27) x2−729

28) x2−784 29) x2−841 30) x2−900

31) x2−961 32) x2−1024 33) x2−1089

34)x2−1156 35) x2−1225 36) x2−1296

37) x2−1369 38) x2−1444 39)x2−1521

40)x2−1600 41) x2−1681 42) x2−1764

43) x2−1849 44) x2−1936 45) x2−2025

46) x2−2116 47) x2−2209 48) x2−2304

49) x2−2401 50) x2−2500 51) x2−2601

52) x2−2704 53) x2−2809 54) x2−2916

References


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196


Company.

Difference of Perfect Square Solutions

1) ( x+1 )(x−1) 2) ( x−2 )(x+2) 3) ( x−3 )( x+3)

4) ( x−4 )(x+4) 5) ( x−5 )(x+5) 6) ( x−6 )(x+6)

7) ( x+7 )(x−7) 8) ( x+8 )(x−8) 9) ( x+9 )(x−9)

10) ( x−10 )(x+10) 11) (x−11)(x+11) 12) ( x−12 )(x+12)

13) (x−13)(x−13) 14) (x+14)(x−14) 15) (x−15)(x+15)

16) (x−16)(x+16) 17) (x+17)(x−17) 18) (x+18)(x−18)

19) (x+19)(x−19) 20) (x−20)(x+20) 21) (x+21)( x−21)

22) (x−22)(x+22) 23) (x−23)(x+23) 24) (x−24 )( x+24 )

25) (x−25)(x+25) 26) (x−26)(x+26) 27) (x+27)(x−27)

28) (x+28)(x−28) 29) (x+29)(x−29) 30) (x−30)(x+30)

31) (x+31)(x−31) 32) (x−32)(x+32) 33) (x+33)(x−33)

34) (x+34)(x−34) 35) (x+35)(x−35) 36) (x−36)(x+36)

37) ( x−37 )(x+37) 38) (x+38)(x−38) 39) (x−39)(x+39)

40) (x+40)(x−40) 41) (x−41)(x+41) 42) (x+42)(x−42)

43) (x+43)(x−43) 44) (x−44)(x+44) 45) (x−45)(x+45)

46) (x+46)(x−46) 47) (x−47)(x+47) 48) (x+48)(x−48)

49) (x−49)( x+49) 50) (x+50)(x−50) 51) (x+51)(x−51)

52) (x−52)(x+52) 53) (x+53)(x−53) 54) (x−54)(x+54)

References


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197


Company.

Additional Difference of Perfect Square Problems with Detailed Solutions

Difference of Perfect Squares Formula:F2−L2=(F−L)(F+ L)

F=first term

L=last term

The following steps will explain in detail how to factor correctly a difference of

perfect squares.

Ask the following questions first when confronted with factoring of any type:


the GCF.






trinomial.


Example:x2−4

For this problem, only two terms are present, which is considered a binomial.

There is a difference sign positioned between the two terms and the x variable is

198

squared. In addition, the number 4 is considered a perfect square. This problem is

called a difference of perfect squares.

Decompose, or factor the variables and numbers separately into:

x∗x−2∗2

This can be rewritten in a more compact form as:

( x )2−(2 )2

The first term is x and the last term is2. Label more distinctly what the terms are

in order to avoid confusion.

F=x

L=2

Using the difference of perfect squares formula, insert the terms.

(x−2)(x+2) Final answer

199


F=first term

L=last t erm


perfect squares.



the GCF.






trinomial.


Example:x2−169

For this problem, only two terms are present, which is considered a binomial,

there is a difference sign positioned between the two terms and the x variable is

squared. In addition, the number 169 is considered a perfect square. This problem is

called a difference of perfect squares.


x∗x−13∗13

200


( x )2−(13 )2

The first term is x and the last term is13. Label more distinctly what the terms

are in order to avoid confusion.

F=x

L=13

Using the difference of perfect squares formula, insert the proper terms.

( x−13 )(x+13)Final answer

201


F=first term

L=last term


perfect square type problem.



the GCF.






trinomial.


Example:4 a2−49c2


there is a difference sign positioned between the two terms and the a∧c variables are

squared. In addition, the numbers 4∧49 are considered perfect squares. This problem

is called a difference of perfect squares.


2a∗2a−7c∗7c

202


(2a )2− (7 c )2

The first term is 2a and the last term is7c. Label more distinctly what the terms

are in order to avoid confusion.

F=2a

L=7 c


(2a−7c)(2a+7 c) Final answer

203


F=first term

L=last term


perfect squares.



the GCF.






trinomial.


Example:14x2− 1

169z2


there is a difference sign positioned between the two terms and the x∧z variables are

squared. In addition, the numbers14∧1

169 are considered perfect squares, even though

fractions are present. This problem is called a difference of perfect squares.


204

12x∗1

2x− 1

13z∗113

z


( 12x)

2

−( 113z )

2

The first term is 12x and the last term is

113z. Label more distinctly what the

terms are in order to avoid confusion.

F=12x

L= 113z


( 12x− 1

13z)( 1

2x+ 1

13z) Final Answer.

205


F=first term

L=last term


perfect squares.



the GCF.






trinomial.


Example:c4−16


there is a difference sign positioned between the two terms and the exponent

associated with the c variable is a multiple of 2. In addition, the number 16 is considered

a perfect square. This problem is called a difference of perfect squares.


c2∗c2−4∗4

206


(c2 )2− (4 )2

The first term is c2 and the last term is4. Label more distinctly what the terms

are in order to avoid confusion, proceed as follows:

F=c2

L=4


(c2+4 )(c2−4 )

Not quite done yet, because the problem is still not factored completely. The

sum of perfect squares(c2+4 ) on the left most side of problem is not factorable any

further, but the difference of squares(c2−4 ) on the right most side of problem is.

Factor c2−4as follows:

c∗c−2∗2


(c )2−(2 )2

The first term is c and the last term is2. Label more distinctly what the terms are

in order to avoid confusion.

F=c

L=2


(c+2)(c−2)

At this time, combine the sum of squares and the difference of squares into one

completed problem as follows:

207

(c2+4 ) (c−2 ) (c+2 )

References


Boston: Pearson.






Company.

End of Section

208

Sum of Squares

1) x2+1 2)x2+4 3) x2+9

4) x2+16 5) x2+25 6) x2+36

7¿ x2+49 8) x2+64 9) x2+81

10) x2+100 11) x2+121 12) x2+144

13) x2+169 14) x2+196 15) z2+225

16) w2+256 17) y2+289 18) x2+324

19) f 2+361 20) x2+400 21) x2+441

22) z2+484 23) y2+529 24) w2+576

25) g2+625 26) f 2+676 27) x2+729

28) x2+784 29) z2+841 30) x2+900

31) x2+961 32) z2+1024 33) w2+1089

34) a2+1156 35) b2+1225 36) y2+1296

37) g2+1369 38) z2+1444 39) w2+1521

40) x2+1600 41) t 2+1681 42) k 2+1764

43) d2+1849 44) f 2+1936 45)s2+2025

46) v2+2116 47) c2+2209 48) h2+2304

49) x2+2401 50) a2+2500 51) z2+2601

52) m2+2704 53) k 2+2809 54) b2+2916

References


Boston: Pearson.





209


Company.

Sum of Square Solutions

College Algebra Students Only1) x=± i 2) x=±2i 3) x=±3 i

4) x=±4 i 5) x=±5 i 6) x=±6 i

7) x=±7 i 8) x=±8 i 9) x=±9 i

10) x=±10 i 11) x=±11 i 12) x=±12i

13) x=±13 i 14) x=±14 i 15) z=±15 i

16) w=±16 i 17) y=±17 i 18) x=±18 i

19) f=±19i 20) x=±20 i 21) x=±21i

22) z=±22i 23) y=±23 i 24) w=±24 i

25) g=±25 i 26) f=±26 i 27 x=±27 i

28) x=±28 i 29) z=±29 i 30) x=±30 i

31) x=±31i 32) z=±32i 33) w=±33 i

34) a=±34 i 35) b=±35i 36) y=±36 i

37) g=±37 i 38) z=±38 i 39) w=±39 i

40) x=±40 i 41) t=±41 i 42) k=±42 i

43) d=±43i 44) f=±44 i 45) s=±45i

46) v=±46 i 47) c=±47i 48) h=±48 i

49) x=±49 i 50) a=±50i 51) z=±51i

52) m=±52i 53) k=±53i 54) b=±54 i

210

Additional Sum of Perfect Square Problems with Detailed Solutions

∑ of Perfect SquareFormula : A2+B2

AlternativeFormula :F2+L2


The most important point to remember when dealing with sum of perfect squares

is the binomial is considered non-factorable. The following steps will discuss in detail

how to solve for a certain variable when confronted with a sum of squares.

Example:x2+16=0

Recognize the problem as having two terms, which is considered a binomial, a

plus sign is positioned between the two terms, the variablex is squared and the number

16 is a perfect square. This leads to the assertion that a sum of perfect squares is

present. Proceed as follows:


1)x2+16=0 Set equation equal to zero.

2)x2+16−16=0−16 Subtract 16 from both sides.

3)x2=−16

4)√ x2=±√−16 Take the square root of both sides.

. 5)x=±√−16

6)x=±√16∗√−1 Decompose term separately.

Remember √−1=i.

7)x=±4 i Final answer.


211






Example:x2+144=0


plus sign is positioned between the two terms, the variablex is squared and the number




1)x2+144=0 Set equation equal to zero.

2)x2+144−144=0−144 Subtract 144 from both sides.

3)x2=−144

4)√ x2=±√−144 Take the square root of both sides.

5)x=±√−144

6)x=±√144∗√−1 Decompose terms separately.

Remember √−1=i.

7)x=±12i Final answer.



212





Example:z2+361=0


plus sign is positioned between the two terms, the variablez is squared and the number




1)z2+361=0 Set equation equal to zero.

2)z2+361−361=0−361 Subtract 361 from both sides.

3)z2=−361

4)√ z2=±√−361 Take the square root of both sides.

5)z=±√−361

6)z=±√361∗√−1 Decompose terms separately.

Remember √−1=i.

7)z=±19 i Final answer.

213

References


Boston: Pearson.






Company.

End of Section

214

Sum of Cubes

1) x3+1 2) 27 x3+64

3) x3+8 4) z3+125

5) x3+27 6) 8 z3+27 x3

7) x3+64 8) z3+216 t3

9) x3+125 10) 64+a3

11) x3+216 12) 8c3+x3

13) x3+343 14) 27+q3

15) x3+512 16) 1000d3+1331

17) x3+729 18) x6+8

19) x3+1000 20) a3+c3

21) x3+1331 22) 8a3c6+27 z3

23) x3+1728 24) 729+c6 z6

25) x3+2197 26) 1000+x9

27) x3+2744 28) m3+4096 x3

29) x3+3375 30) 1000+729 c3

31) x3+4096 32) h6+q6

33) x3+4913 34) 1331 c3+2744

35) x3+5832 36) 125+a3b3c3

37) x3+6859 38) 3375+c3 k3

39) x3+8000 40) 512x3+z6

41) x3+9261 42) 343 g3+125

43) x3+10648 44) 1+8000h3

215

45) x3+12167 46) 8a6+27b6

47) x3+13824 48) 64+ x3 y3 z3

49) x3+15625 50) l3+8000

References


Boston: Pearson.






Company.


216

Sum of Cube Solutions

1) ( x+1 ) (x2−x+1 ) 2) (3 x+4)(9x2−12 x+16)

3) (x+2)( x2−2x+4 ) 4) (z+5)(z2−5 z+25)

5) (x+3)(x2−3 x+9) 6) (2 z+3x )(4 z2−6xz+9x2)

7) (x+4 )( x2−4 x+16) 8) (z+6 t)(z2−6 tz+36 t2)

9) (x+5)(x2−5x+25) 10) (4+a)(16−4a+a2)

11) (x+6)(x2−6 x+36) 12) (2c+x)(4c2−2cx+x2)

13) (x+7)(x2−7 x+49) 14) (3+q)(9−3q+q2)

15) (x+8)(x2−8 x+64) 16) (10d+11)¿)

17) (x+9)(x2−9 x+81) 18) (x2+2)(x4−2 x2+4)

19) (x+10)(x2−10x+100) 20) (a+c)(a2−ac+c2)

21) (x+11)(x2−11 x+121) 22) (2ac2+3 z)(4a2 c4−6ac2 z+9 z2)

23) (x+12)(x2−12 x+144) 24) (9+c2 z2)(81−9c2 z2+c4 z4)

25) (x+13)(x2−13x+169) 26) (10+x3)(100−10 x3+ x6)

27) (x+14)(x2−14 x+196) 28) (m+16 x)(m2−16mx+256 x2)

29) (x+15)(x2−15x+225) 30) (10+9 c)(100−90c+81c2)

31) (x+16)(x2−16 x+256) 32) (h2+q2)(h4−h2q2+q4)

33) (x+17)(x2−17 x+289) 34) (11c+14)(121c2−154 c+196)

35) (x+18)(x2−18x+324) 36) (5+abc)(25−5abc+a2b2c2)

37) (x+19)(x2−19x+361) 38) (15+ck)(225−15 ck+c2 k2)

39) (x+20)(x2−20 x+400) 40) (8+z2)(64−8 z2+z 4)

41) (x+21)( x2−21x+441) 42) (7 g+5)(49g2−35g+25)

217

43) (x+22)( x2−22x+484 ) 44) (1+20h)(1−20h+400h2)

45) (x+23)(x−23 x+529) 46) (2a2+3b2)(4 a4−6a2b2+9b4)

47) (x+24)(x−24 x+576) 48) (4+xyz )(64−4 xyz+x2 y2 z2)

49) (x+25)(x2−25 x+625) 50) (t+20)(t2−20t+400)

References


Boston: Pearson.






Company.

218

Additional Sum of Cube Problems with Detailed Solutions

Sum of Cubes Formula: F3+L3=(F+L )(F2−FL+L2)

F=First term

L=Last term

FL=F∗L(Implied)

The following steps will explain in detail how to factor the following type of

problems.

Example:x3+8

Questions that should be considered every time: “How many terms do I have?”

“Do I have a binomial?” “Do I have a sum of squares, a difference of squares, a sum

of cubes or a difference of cubes?”


sign is inserted between the two terms and the exponent is cubed. This leads to the

assertion that a sum of cubes is present. Decompose, or factor each term separately.

x∗x∗x+2∗2∗2.

Rewrite in a more compact form as:

( x )3+ (2 )3

The term that is inside the first set of parenthesis is considered theFirst. The

term that is inside the second set of parenthesis is considered theLast . Now, using the

sum of cubes formula, insert the terms.

( x+2 )(( x )2−2∗x+(2 )2)

219

One more step needs to be done in order to “clean up” the answer and that is to

factor thex2term and the22term intox∗x and2∗2.

(x+2)( x∗x−2 x+2∗2)

(x+2)( x2−2x+4 )

220


F=First term

L=Last term

FL=F∗L(Implied)

Example:x3+125







x∗x∗x+5∗5∗5

Rewrite in a more compact form as follows:

( x )3+ (5 )3.

The First term is x and the Last term is5. Using the sum of cubes formula, insert

the correct terms:

(x+5)(( x )2−5∗x+(5 )2)



(x+5)(x∗x−5 x+5∗5)

(x+5)(x2−5x+25)

221


F=First term

L=Last term

FL=F∗L(Implied)

Example:a3b3+216 c3







a∗a∗a∗b∗b∗b+6∗6∗6∗c∗c∗c

Rewrite in a more compact form as (ab )3+ (6c )3.

The First term is ab and the Last term is6c. Label what the first term and the

last term are in order to keep from becoming confused.

F=ab L=6 c

Proceed to insert the terms into the sum of cubes formula:

(ab+6c)( (ab )2−6∗a∗b∗c+(6 c )2)


factor the¿term and the(6c )2term intoab∗ab and6c∗6 c.

(ab+6c)(ab∗ab−6abc+6c∗6 c)

(ab+6c)(a2b2−6 abc+36c2) Final Answer.

222


F=First term

L=Last term

FL=F∗L(Implied)

Example:27 x3+ 164z3

Questions should be considered every time: “How many terms do I have?” “Do I

have a binomial?” “Do I have a sum of squares, a difference of squares, a sum of

cubes or a difference of cubes?”




3 x∗3 x∗3 x+ 14z∗14

z∗14

z

Rewrite in a more compact form as(3 x )3+( 14z)

3

. The First term is3 x and the Last

term is14z. Label what the first term and the last term are in order to keep from

becoming confused.

F=3 x L=14z

Proceed to insert terms into sum of cubes formula:

(3 x+ 14z )((3 x )2−3 x∗1

4z+( 1

4z)

2

)

223


factor the¿term and the( 14z)

2

term into3 x∗3 x and14z∗1

4z.

(3 x+ 14 z

)(3 x∗3 x−34xz+ 1

4z∗14

z)

(3 x+ 14z )(9 x2−3

4xz+ 1

16z2)

224


F=First term

L=Last term

FL=F∗L(Implied)

Example:x6+8





sign is inserted between the two terms and the exponent associated with the x term is

raised to a multiple of three. This leads to the assertion that a sum of cubes is present.

Decompose, or factor each term separately.

x2∗x2∗x2+2∗2∗2

Rewrite in a more compact form as(x2 )3+(2 )3.

The First term is x2 and the Last term is2. Label what the first term and the last

term are in order to keep from becoming confused.

F=x2 L=2


(x2+2)(( x2 )2−2∗x2+ (2 )2)



(x2+2)(x2∗x2−2x2+2∗2)

(x2+2)(x4−2 x2+4)

225


F=First term

L=Last term

FL=F∗L(Implied)

Example:343a3+8000 k 3







7a∗7a∗7a+20 k∗20k∗20k

Rewrite in a more compact form as(7a )3+ (20 k )3.

The First term is 7a and the Last term is20k. Label what the first term and the


F=7a L=20k


(7a+20k)( (7a )2−7 a∗20 k+ (20k )2)


factor the(7a )2term and the(20k )2term into7a∗7a and20k∗20k.

(7a+20k)(7a∗7 a−140ak+20k∗20k )

(7a+20k)(49a2−140ak+400 k2)

226

References


Boston: Pearson.






Company.

End of Section

227

Difference of cubes

1) x3−1 2) t 3−1000

3)x3−8 4) p6−8

5) z3−27 6) 27 x3−64 z3

7) x3−64 8) p3d3q3−125

9) q3−125 10) 1−r3d3

11) s3−216 12) h3− j3

13)c3−343 14) 1000−x9

15) v3−512 16) m3−4096 x3

17) m3−729 18) 64−r9

19) x3−1000 20) 27−8 y3

21) a3−1331 22) 8c3−x3

23) d3−1728 24) 3375−g3d3

25) p3−2197 26) x6−8

27) k3−2744 28) z3−8000

29) y3−3375 30) 8a6−27b6

31) c3−4096 32) 512c3−z6

33) x3−4913 34) 216w3−1331h3

35) r3−5832 36) 1−z3

37) h3−6859 38) 3375− y6

39) p3−8000 40) c3−4096

41) b3−9261 42) 216−d6v6

43) j3−10648 44) 216m3−4913

228

45) m3−12167 46) 343b3−1

47) c3−13284 48) x3 y3 z3−343

49) m3−15625 50) 64−z3t 3

References


Boston: Pearson.






Company.


229

Difference of Cube Solutions

1) (x−1)(x2+x+1) 2) (t−10)(t 2+10 t+100)

3) (x−2)(x2+2x+4 ) 4) ( p2−2)( p4+2 p2+4)

5) (z−3)(z2+3 z+9) 6) (3 x−4 z)(9x2+12xz+16 z2)

7) (x−4)(x2+4 x+16) 8) ( pdq−5)( p2d2q2+5 pdq+25)

9) (q−5)(q2+5q+25) 10) (1−rd)(1+rd+r 2d2)

11) (s−6)(s2+6 s+36) 12) (h− j )(h2+hj+ j2)

13) (c−7)(c2+7 c+49) 14) (10−x3)(100+10 x3+ x6)

15) (v−8)(v2+8 v+64) 16) (m−16 x )(m2+16mx+64 x2)

17) (m−9)¿ 18) (4−r3)(16+4 r3+r6)

19) (x−10)(x+10 x+100) 20) (3−2 y )(9+6 y+4 y2)

21) (a−11)(a2+11 a+121) 22) (2c−x )(4 c2+2cx+x2)

23) (d−12)(d2+12d+144) 24) (15−gd )(225+15 gd+g2d2)

25) ( p−13)( p2+13 p+169) 26) (x2−2)(x4+2 x2+4)

27) (k−14)(k2+14 k+196) 28) (z−20)(z2+20 z+400)

29) ( y−15)( y2+15 y+225) 30) (2a2−3b2)(4b4+6a2b2+9b4)

31) (c−16)(c2+16 c+256) 32) (8c−z2)(64c2+8 c z2+z4)

33) (x−17)(x2+17 x+289) 34) (6w−11h)¿

35) (r−18)(r+18 r+324) 36) (1−z )(1+z+z2)

37) (h−19)(h2+19h+361) 38) (15− y2 )(225+15 y2+ y4)

39) ( p−20)( p+20 p+400) 40) (c−16)(c2+16 c+256)

41) (b−21)(b2+21b+441) 42) (6−d2 v2)(36+6d2v2+d4 v 4)

230

43) ( j−22)( j2+22 j+484) 44) (6m−17)(36m2+102m+289)

45) (m−23)(m2+23m+529) 46) (7b−1)(49b2+7b+1)

47) (c−24)(c2+24c+576) 48) (xyz−7)(x2 y2 z2+7 xyz+49)

49) (m−25)(m2+25m+625) 50) (4− zt)(16+4 zt+z2t 2)

References


Boston: Pearson.






Company.

231

Additional Difference of Cube Problems with Detailed Solutions

Difference of Cubes Formula:F3−L3=(F−L)(F2+FL+L2)

F=First term

L=Last term

FL=F∗L(implied)

Example: x3−64

Ask the following questions when faced with any aspect of factoring: “How many

terms do I have?” “Do I have two terms, three terms or four terms?” “Do I have a

difference of squares, a sum of squares, a difference of cubes or a sum of

cubes?”

For the above problem, two terms are present, which implies a binomial. A minus

sign is positioned between the two terms, and the exponent is cubed. These three

conditions assert a difference of cubes is present. The following steps will show how

to factor correctly the above problem.

Decompose, or factor each term separately into the following:

x∗x∗x−4∗4∗4

This can be rewritten in a more compact form as( x )3−( 4 )3.

The First term is x and the Last term is4. Label what the first term and the last


F=x L=4

Using the formula for difference of cubes, insert the terms:

( x−4 )( (x )2+4∗x+( 4 )2)

232



(x−4)(x∗x+4 x+4∗4 )

(x−4)(x2+4 x+16)

233


F=First term

L=Last term

FL=F∗L(implied)

Example: z3−8000

Always ask the following questions when faced with the aspect of any factoring:

“How many terms do I have?” “Do I have two terms, three terms or four terms?” “Do I

have a difference of squares, a sum of squares, a difference of cubes or sum of

cubes?”






z∗z∗z−20∗20∗20

This can be rewritten in a more compact form as( z )3−(20 )3.

The First term is z and the Last term is20. Label what the first term and the last


F=z L=20

Using the formula for difference of cubes, insert the terms.

( z−20 )(( z )2+20∗z+(20 )2)

234


factor thez2term and the202term intoz∗z and20∗20.

(z−20)(z∗z+20 z+20∗20)

(z−20)(z2+20 z+400)

235


F=First term

L=Last term

FL=F∗L(implied)

Example:a3b3c3−x3 y3 z3



have a difference of squares, a sum of squares, a difference of cubes or a sum of

cubes?”






a∗a∗a∗b∗b∗b∗c∗c∗c−x∗x∗x∗y∗y∗y∗z∗z∗z

This can be rewritten in a more compact form as(abc )3−( xyz )3.

The First term is abc and the Last term isxyz. Label what the first term and the


F=abc L=xyz


(abc−xyz )( (abc )2+abc∗xyz+( xyz )2)


factor the(abc )2term and the(xyz )2term intoabc∗abc andxyz∗xyz.

236

(abc−xyz )(abc∗abc+abcxyz+xyz∗xyz )

(abc−xyz )(a2b2c2+abcxyz+x2 y2 z2)

237


F=First term

L=Last term

FL=F∗L(implied)

Example: 1

1000x3−64 k3



have a difference of squares, a sum of squares, a difference of cubes or sum of

cubes?”






110

x∗110

x∗110

x−4k∗4 k∗4k

This can be rewritten in a more compact form as( 110x )

3

−(4 k )3.

The First term is 1

10x and the Last term is4 k. Label what the first term and the


F= 110x L=4k


238

( 110x−4k )(( 1

10x)

2

+ 110x∗4k+( 4k )2)


factor the( 110x)

2

term and the(4k )2term into1

10x∗110

x and4 k∗4 k.

( 110x−4k )( 1

10x∗110

x+ 25kx+4 k∗4k )

( 110x−4k )( 1

100x2+ 2

5kx+16k2)

239


F=First term

L=Last term

FL=F∗L(implied)

Example: x3−262144




cubes?”






x∗x∗x−64∗64∗64

This can be rewritten in a more compact form as( x )3−(64 )3.

The First term is x and the Last term is64. Label what the first term and the last


F=x L=64


( x−64 )(( x )2+64∗x+ (64 )2)


factor the(x)2term and the(64)2term intox∗x and64∗64.

240

(x−64 )(x∗x+64 x+64∗64)

(x−64 )(x2+64 x+4096)

241


F=First term

L=Last term

FL=F∗L(implied)

Example:1000−a3




cubes?”






10∗10∗10−a∗a∗a

This can be rewritten in a more compact form as(10 )3− (a )3.

The First term is 10 and the Last term isa. Label what the first term and the last

term are in order to keep from becoming confused

F=10 L=a


(10−a ) ¿

242


factor the(10)2term and the(a)2term into10∗10 anda∗a.

(10−a)(10∗10+10a+a∗a)

(10−a)(100+10a+a2)

References


Boston: Pearson.






Company.

End of Section

243

Monic Trinomials

x2+bx+c

1) x2−5 x−14 2) x2−19 x+78

3) x2+9 x+14 4) y2+23 y+132

5) x2−2 x−63 6) n2−7n−114

7) x2−4 x−45 8) a2−17a+70

9) x2+14 x+45 10) c2−3c−180

11) z2−40 z−84 12) h2−5h+4

13) z2−44 z+84 14) f 2−9 f+8

15) c2+13c+22 16) k 2+22k+117

17) d2+88d−180 18) h2+29h+198

19) s2+93 s+270 20) d2−5d−84

21) z2+40 z−225 22) z2−17 z+16

23) t 2−121t−246 24) k 2−8k−105

25) x2−101 x+480 26) h2+7h−198

27) t 2+98 t+97 28) b2+26b+153

29) v2+40v+111 30) x2−3 x−154

31) f 2+63 f +392 32) t 2−8 t−180

33) k 2−49k−392 34) c2−23c+130

35) g2−724 g+2880 36) x2−31 x+220

37) a2−716a−2880 38) m2−3m−304

39) s2−260 s−789 40) b2−22b+112

244

References


Boston: Pearson.






Company.


245

Monic Trinomial Solutions

x2+bx+c

1) ( x−7 )(x+2) 2) ( x−6 ) ( x−13 )

3) ( x+7 )(x+2) 4) ( y+11) ( y+12 )

5) ( x−9 )(x+7) 6) (n+6 ) (n−19 )

7) ( x−9 )(x+5) 8) (a−10 ) (a−7 )

9) ( x+9 )(x+5) 10) (c+12 ) (c−15 )

11) ( z−42 )(z+2) 12) (h−4 ) (h−1 )

13) ( z−42 )(z−2) 14) ( f−8 ) ( f−1 )

15) (c+11)(c+2) 16) (k+13 ) ( k+9 )

17) (d+90 )(d−2) 18) (h+11 ) (h+18 )

19) ( s+90 )(s+3) 20) (d+7 ) (d−12 )

21) ( z+45 )(z−5) 22) ( z−16 ) ( z−1 )

23) ( t+2 )(t−123) 24) (k+7 ) (k−15 )

25) ( x−5 )(x−96) 26) (h−11) (h+18 )

27) ( t+97 ) ( t+1 ) 28) (b+9 ) (b+17 )

29) ( v+3 ) ( v+37 ) 30) ( x+11)(x−14 )

31) ( f +56 )( f +7) 32) (t−18 )(t+10)

33)(k−56 )(k+7) 34) (c−10 ) ( c−13 )

35) (g−4 ) (g−720 ) 36) ( x−11 ) ( x−20 )

37) (a+4 ) (a−720 ) 38) (m+16 ) (m−19 )

39) ( s−263 ) (s+3 ) 40)(b−14 )(b−8)

246

References


Boston: Pearson.






Company.

247

Additional Monic Trinomial Problems with Detailed Solutions

x2+bx+c



the GCF.






trinomial.


Example:x2+16x+63

For the above problem, there is no GCF and three terms are present, which

leads to the assertion that a monic trinomial is present. A monic trinomial contains a

coefficient of 1 in front of the x2 term.

The overarching strategy of monic trinomials is to focus on the last term of the

problem, considering paired factors of that certain last number. In other words, “What

two numbers multiplied together will produce or yield the last number”? Do not be too

overly concerned with the signs of the above problem at the present time, the signs will

"fall into place" at a later time. Proceed as follows:

248


1)x2+16x+63 Recopy problem.

List all paired factors of 63 63*121*39*7

Which pair of factors if added or subtracted, is the result16? The pair of factors

whose product is63, but sum is16are9∧7. At this point, the trinomial is factorable.

Proceed as follows:

2)( x )(x) Begin with parenthesis, and then insert

variables.

3)( x 9 )(x7) Insert correct numbers, noting it does

not matter if 9∨7are reversed!

4)( x+9 )(x+7) If original problem has two plus signs,

then both signs will be positive.

5)( x+9 )(x+7) Final answer.

249



the GCF.






trinomial.


Example:x2+32 x+135









250


1)x2+32x+135 Recopy problem.

List all paired factors of 135 135*145*327*515*9


whose product is135, but sum to 32 are27∧5. At this point, the trinomial is factorable.

Proceed as follows:


variables.

3)( x27 )(x5) Fill in correct numbers, noting that it

does not matter if 27∧5 are

reversed!

4)( x+27 )(x+5) If original problem has two plus signs,

then both signs will be positive.

5)( x+27 )(x+5) Final answer

Notice how the signs are placed last in the process! In addition, don’t worry so

much about how the numbers are placed. If numbers were swapped by placing the 5

first, then the27, no harm done since both positive signs are present in original problem!

251



the GCF.






trinomial.


Example:z2−30 z−99



coefficient of 1 in front of the z2 term.


problem, considering factors of that certain last number. In other words, “What two

numbers multiplied together will produce or yield the last number”? Do not be too overly

concerned with the signs of the above problem at the present time, the signs will "fall

into place" at a later time. Proceed as follows:

252


1)z2−30 z−99 Recopy problem.

List all paired factors of 99 99*133*311*9


whose product is99, but difference is 30 are 33∧3. If this cannot be seen mentally, then

jot down on scratch paper33∧3. The product of these two numbers is99, the sum of

these two numbers is36, but the difference of these two numbers is30.

Look at the middle term and see that the difference matches correctly. Again,

don't be overly concerned about the positioning of the signs; they will "fall into place", so

to speak at a later time. At this point, the trinomial is factorable. Proceed as follows:


2)( z )(z) Begin with parenthesis, and then insert

variables.

3)( z3 )(z33) Fill in correct numbers.

If original problem has two negative signs inserted, then one of the signs will be

negative and the other one will be positive. The sign that is in front of the middle term of

the original problem dictates where it is positioned; therefore it is inserted in front of the

larger of the two numbers. "33 is much larger than 3", so the minus sign falls in front of

the33.

4)( z+3 )(z−33) Final answer.

253

Notice how the signs are placed last in the process! Placement of the signs in

the last part of process is critical in order to keep confusion and stress level to a

minimum!

254



the GCF.






trinomial.


Example:x2−85 x+166





equation, considering paired factors of that certain last number. In other words, “What




255


1)x2−85 x+166 Recopy problem.

List all paired factors of 166 166*183*2


whose product is166, but whose sum is 85 is83∧2. If this cannot readily be seen

mentally, then jot down on scratch paper83∧2. The product of these two numbers is 166

.The sum of these two numbers is85, but the difference of these two numbers is81.

Look at the middle term and see that the sum matches correctly. Again, don't be

overly concerned about the positioning of the signs, they will "fall into place", so to

speak at a later time. At this point, the trinomial is factorable. Proceed as follows:


variables.

3)( x2 )( x83) Fill in correct numbers, noting that it

does not matter if 2∧83 are

reversed!

"When last term is positive and middle term negative of the original problem,

then the signs will be: minus, minus."

4)( x−2 )(x−83) Final Answer.

Notice how the signs are placed last in the process! Placement of the signs in

the last part of process is critical in order to keep confusion and stress level to a

minimum!

256



the GCF.






trinomial.


Example:a2+21a−72



coefficient of 1 in front of the a2 term.






257


1)a2+21a−72 Recopy problem.

List all paired factors of 72 72*136*224*318*412*6 9*8


whose product is72, but whose difference is 21 is24∧3. If this cannot readily be seen

mentally, then jot down on scratch paper 24∧3.The product of these two numbers is72,

the sum of these two numbers is27, but the difference of these two numbers is21. Look

at the middle term and see that the difference matches correctly. Again, don't be overly

concerned about the positioning of the signs; they will "fall into place" so to speak at a

later time. At this point, the trinomial is factorable. Proceed as follows:


2)(a )(a) Begin with parenthesis, and then insert

variables.

3)(a3 )(a24) Fill in correct numbers.

If original problem has a minus and a plus sign, then one of the signs will be

negative and the other will be positive. The sign that is in front of the middle term of the

original problem dictates where it is positioned; therefore it is inserted in front of the

larger of the two numbers.24is much larger than 3, so the plus sign falls in front of the

24.

4)(a−3 )(a+24) Final answer

Notice how the signs are placed last in the factoring process!

258

References


Boston: Pearson.






Company.

End of Section

259


a x2+bx+c a≠1 ,0

1) 2 x2+15 x+28 2)35 g2+99 g−72

3) 10 x2−7 x−12 4) 33 z2−4 z−37

5) 8a2−26a+15 6) 15q2+56q+49

7)2 x2+x−6 8) 15b2−14 b−49

9)3 x2−11 x+6 10) 19h2−80h−99

11)4 a2+9a−9 12) 19 j2+24 j−99

13)33v2+70v+37 14) 81a2−181a+100

15)4 c2−12 c+5 16) 81h2+225h+100

17) 4w2−13w+3 18) 25 r2+88 r+39

19) 5 y2−12 y−9 20) 25 t 2−14 t−39

21) 2 x2−5 x+3 22) 42k 2+11 k−31

23) 3 y2+8 y+4 24) 42 l2+193l+31

25) 6 t 2−17 t+5 26) 27 x2−51xz+22 z2

27)21a2+5ac−6 c2 28) 48 d2−19d−67

29) 3a2−10ax+3 x2 30)9 j2−56 j+47

References


Boston: Pearson.




260



Company.


261

Non-Monic Trinomial Solutions

a x2+bx+c a≠1

1) (2 x+7 ) ( x+4 ) 2) (7 g+24 )(5g−3)

3) (5 x+4 ) (2 x−3 ) 4) (33 z−37 )( z+1)

5) (4 a−3 ) (2a−5 ) 6) (5q+7 )(3q+7)

7) (2 x−3 ) ( x+2 ) 8) (5b+7 ) (3b−7 )

9) (3 x−2 ) ( x−3 ) 10) (19h−99 ) (h+1 )

11) (4 a−3 )(a+3) 12) (19 j−33 )( j+3)

13) (33v+37 ) ( v+1 ) 14) (81a−100 )(a−1)

15) (2c−1 ) (2c−5 ) 16) (9h+5 )(9h+20)

17) (4w−1 ) (w−3 ) 18) (25 r+13 )(r+3)

19) (5 y+3 ) ( y−3 ) 20) (25 t−39 )(t+1)

21) (2 x−3 ) ( x−1 ) 22) (42k−31 )(k+1)

23) (3 y+2 ) ( y+2 ) 24)(7 l+31 )(6 l+1)

25) (3 t−1 ) (2t−5 ) 26) (9 x−11 z )(3 x−2 z)

27) (7a−3c ) (3 a+2c ) 28)(48 d−67 )(d+1)

29) (3a−x ) (a−3 x ) 30) (9 j−47 )( j−1)

References


Boston: Pearson.




262



Company.

263

Additional Non-Monic Trinomial Problems with Detailed Solutions

a x2+bx+c a≠1



the GCF.






trinomial.


Example:2 x2+15 x+28

The above problem is considered a non-monic trinomial due to the presence of a

coefficient other than 1 in front of the x2 term. The coefficient or number in front of the

first term is2. The following steps will detail how to factor properly this problem.


1)2 x2+15 x+28 Recopy problem, noting mentally the 15x term.

2) Decompose first and last term into all factors.

264

2x

x

28

1

14

2

The overarching strategy for non-monic trinomials is to find two products and if

their sum or difference equals the middle term, then the trinomial is factorable. The

ability to decompose the first term and the last term into smaller factors is for fast

mental computation of the two products. This can be time consuming at first, but after

practice, the process becomes more efficient.


First product :2 x∗28=56 x

Second product : x∗1=x

∑ :56 x+x=57 x

Difference :56 x−x=55 x

At this time, we are attempting to find a sum or difference that equals 15 x if the

two products are added or subtracted. The two products are56 x∧x. It is apparent that

the sum or difference of 15 x has not been achieved. Try using a different combination of

factors by implementing a criss-cross pattern.

265

7

4

2x

x

28

1

2x

x

28

1


Second product : x∗28=28 x

∑ :2 x+28 x=30 x

Difference :28 x−2x=26 x

At this time the number 15 x is still not present. Again use different combinations

of factors for 28.

3)



∑ :28 x+2 x=30 x

Difference :28 x−2x=26 x

At this time, the number 15 x has still not been found by adding or subtracting the

two products. The two products are28 x∧2 x. Use different combinations of factors by

trying a criss-cross pattern.



∑ :14 x+4 x=18 x

Difference :14 x−4 x=10 x

266

2x

x

14

2

2x

x

14

2

This process does take time upon first exposure, but with practice, the time

attempting to find the middle term decreases dramatically. The two products are

4 x∧14 x. Their sum or difference does not equal15 x, though.

Again use different combinations of factors. Now use the last set of factors for28.

4)



∑ :14 x+4 x=18 x


This process does take time upon first exposure, but with practice, the time

attempting to find the middle term decreases dramatically. The two products are again

14 x∧4 x. Their sum or difference does not equal15 x, though.

Now, attempt a different combination of factors by trying a criss-cross pattern.

First product :2 x∗4=8x


∑ : 8x+7 x=15 x

Difference :8 x−7 x=x

267

2x

x

7

4

2x

x

7

4

BINGO!

Notice, the criss cross pattern works. At this time, 15 x has now been found.

Finally!!!! This non-monic trinomial is factorable. Proceed as follows:

5) ()() "Open, close, open, close".

6) (2 x)(x ) Insert 2 x∧x from left column.

7) (2 x)(x 4) Remember that the pair 2 x∧4 are

mated together to represent OUTER

portion of the FOIL process.

8) (2 x7)(x 4) Remember that the pair x∧7 are

mated together to represent the INNER


At this time, no signs have been inserted, to avoid confusion. If two positive signs

are present in the original problem, then two positive signs will be inserted in the

factored problem.

9)(2 x+7)(x+4) Final answer.

This process is time consuming for the beginner and maybe a bit confusing upon

initial exposure, but after practice, the process becomes extremely efficient. In addition,

dealing with smaller factors and products helps with mental computations. The overall

reason for placement of arrows is to “keep track” of pairing of numbers.

268



the GCF.






trinomial.


Example:3 x2−11 x+6


coefficient other than1 in front of the x2 term. The coefficient or number in front of the

first term is3. The following steps will detail how to factor properly this problem.


1)3 x2−11 x+6 Recopy problem, noting mentally the

11x term.

2) Decompose first and last term into

all factors.

269

3x

x

6

1

3

2

3) The overarching strategy for non-monic trinomials is to find two products and if their

sum or difference equals the middle term, then the trinomial is factorable. This can be

time consuming at first, but after practice, the process becomes extremely efficient.




∑ :18 x+x=19 x


We are attempting to find a sum or difference that equals 11 x if the two products

are added or subtracted. The two products are18 x∧x. Their sum or difference does not

equal11 x, though. Try using a different combination of factors by attempting a criss-

cross pattern.



∑ :3 x+6 x=9 x


270

3x

x

6

1

3x

x

6

1

Notice, the criss-cross pattern works and is allowed. We are attempting to find a

sum or difference of 11 x by adding or subtracting the two products. The two products

are3 x and6 x. Their sum or difference does not equal11 x, though. Attempt again, but

using different combinations of factors for6.



∑ : 9x+2 x=11 x

Dif ference :9 x−2x=7x

BINGO!

A sum of 11x has indeed been reached by adding the two products. The two

products are9 x∧2 x. This non-monic trinomial is factorable. Further steps are required

to finish problem.


6) (3 x)(x ) Insert 3 x∧x from the first column.


mated together to represent the OUTER


271

3x

x

3

2

8) (3 x2)(x3) Remember that the pair x∧2 are



Again, the signs are not inserted until the very last part of the factoring process in

order to avoid confusion and minimize stress levels. If the last term is positive and

middle term is negative in original problem, then both signs will be negative. Proceed

as follows:

9) (3 x−2)(x−3) Final answer.

The overall reason for placement of arrows is to “keep track” of pairing of

numbers. One easy trick that could be implemented is the use of Popsicle® sticks as

the “arrows”, or use two pencils as a guide.

272



the GCF.






trinomial.


Example:45 x2+122 x+77


coefficient other than1 in front of the x2 term. The coefficient or number that is in front of

first term is45.

Notice how large the first term’s coefficient is, as well as last term. These “big

numbers” can lead to increased stress levels. The following steps will detail how to

factor properly this problem and deal with the large numbers more effectively.


1) 45 x2+122 x+77 Recopy problem, noting mentally 122 x term.


all factors.

273

45x

x

77

1



time consuming at first and maybe a bit awkward, but after practice, the process

becomes extremely efficient.


First product : 45x∗77=3465 x


∑ :3465 x+x=3466 x


We are attempting to find a sum or difference of 122 x by adding or subtracting

the two products. The two products are3465 x∧x. Their sum or difference does not

equal122 x, though. Try again, but using different combinations of factors.

274

15x

3x

11

7

9x

5x

45x

x

77

1

45x

x

77

1

First product : 45x∗1=45 x


∑ : 45 x+77 x=122 x


BINGO!

A sum of 122 x has been reached by adding the two products. This non-monic

trinomial is factorable. Further steps are required to finish problem.


6) (45x )(x) Insert 45 x∧x from the first column.

7) (45 x ) ( x1 ) Remember that the pair 45 x∧1 are



8) (45x 77)(x1) Remember that the pair x∧77 are



9) Again, the signs are not inserted until the very last part of the factoring process in

order to avoid confusion and minimize stress levels.

If both signs are positive from the original problem, then the signs inserted will

also be positive. It is much easier to insert signs at end of factoring process.

10) (45 x+77)(x+1) Final answer.

275



the GCF.






trinomial.


Example:33 x2+4 x−37


coefficient other than1 in front of the x2 term. The coefficient or number in front of the

first term is33. Notice how large the first term’s coefficient is, as well as last term. These

“big numbers” can lead to increased stress levels. The following steps will detail how to



1) 33 x2+4 x−37 Recopy problem, noting mentally the 4 x

term.

276


factors.


sum or difference equals the middle term, then the trinomial is factorable.

This can be time consuming at first, and maybe a bit awkward, but after practice,

the process becomes extremely efficient.




∑ :1221 x+ x=1222 x


We are attempting to find a sum or difference of 4 x by adding or subtracting the

two products. The two products are1221 x∧x. These two numbers do not add or subtract

to attain4 x, though. Try again, but using different combinations of factors.

277

33x

x

37

1

11x

3x

33x

x

37

1

33x

x

37

1



∑ :33 x+37 x=70 x


BINGO!

A difference of 4 x has been reached by subtracting the two products. This non-

monic trinomial is factorable. Further steps are required to finish problem.

5) (33 x)(x ) Insert 33 x∧x from left column.




7) (33 x37)(x 1) Remember that the pair x∧37 are




order to avoid confusion and minimize stress levels. If signs are positive and negative

from the original problem, then the signs inserted into factored problem will also be

positive and negative.

The question that arises is “Where to put the signs, and why?” It is easier to look

at the two products that were produced in the earlier steps; the larger product “takes”

the sign of the middle coefficient from the original problem. For the above problem,

33 x is considered the OUTER product and 37 x is considered the INNER product, so the

positive sign must be placed in front of the 37 and the negative sign placed in front of

278

the 1 in order to ensure when you subtract the two products,37 x−33 x=4 x. It is much

easier to insert signs at end of factoring process.

9) (33 x+37)(x−1) Final answer.

279



the GCF.






trinomial.


Example: 21 x2−65x+24


coefficient other than1in front of the x2 term. The coefficient or number in front of the first

term is21. Notice how large the first term’s coefficient is, as well as last term. These “big

numbers” can lead to increased stress levels. The following steps will detail how to



1) 21 x2−65x+24 Recopy problem, noting mentally the

65 x term.


all factors.

280

7x

3x

8

3



time consuming at first, but after practice, the process becomes extremely efficient.



Second product :3x∗3=9x

∑ :56 x+9 x=65 x


BINGO!

A sum of 65 x has been reached by adding the two products. The two products

are 56 x∧9 x.This non-monic trinomial is factorable. Further steps are required to finish

problem.

281

21x

x

12

2

6

4

24

1

7x

3x

8

3


6) (7 x)(3x ) Insert 7 x∧3 x from left column.

7) (7 x)(3x 8) Remember that the pair 7 x∧8 are



8) (7 x3)(3 x 8) Remember that the pair 3 x∧3 are




order to avoid confusion and minimize stress levels. Use the following phrase, "When

last term is positive and middle term is negative from the original problem, then both

signs will be negative".

10) (7 x−3)(3 x−8) Final answer.

References


Boston: Pearson.






Company.

282

End of section

283

"Solve by Factoring” Problems

Solve the following equations:

1) 2 x2=13x+7 2) 2 x3−7 x2+6 x=0

3) 3 x2=7 x 4) 3u3−2u2−8u=0

5) ( x−3 ) ( x−2 )=12 6) 4 n3−4n2=3n

7) x (x+5 )−10=8 x 8) 4m3+5m2=6m

9) 2 x3+5 x2=3 x 10) 3 x2+8 x−3=0

11) x2−5 x+6=0 12) 2 x2=7 x−6

13) x2+7x+10=0 14) x2=3 x

15) x2+ x−2=4 16) 2x5

+ 1115

= 23 x

17) t 2−6 t−7=9 18) 3 xx−4

− 22x2−4 x

=5x

19) 2m2−7m=−6 20) 2xx−3

− 6 xx2−9

= xx+3

21) 3 x2+11 x=−6 22) 116

=2x3

− 12x

23) (g−1 ) (g+2 )=10 24) 4 x

2x2+5 x−3+ 3x2+3x

= 12x2−x

25) ( p−5 ) ( p+2 )=−12 26) 35−5

8= x

20

27) (2 x−1 ) (x+1 )=5 28) 13+ 5

6=1x

29) (3w−2 ) (w−2 )=7 30) 16+ 1

8=1t

31) (4 x+3 ) ( x−2 )=0 32) x−8x+3

=14

284

33) (5n+7 ) (2n−3 )=0 34) x+1

3−1= x−1

2

35) 6 x2−9 x+3=2 x 36)x+5x=−6

37) 4 x2−9 x+5=3 x 38) x6−6x=0

39) x2=9 40) 2x+1

= 1x−2

41) 9 x2=16 42) 9n−13n−2

= 53n−2

+3

43) m (m+3 )=18 44) 2

3b−1= 6b

3b−1−2

45) −t (7−t )=18 46) 5

c2+2c=3c− 7

2c

47) 5−3a

a2+4 a+3−2a+2a+3

=3−aa+1 48)

3a−5

−23= 1−8a

3a−15

49) 1x−1

+x−5=5x−4x−1

−6 50) 4 t−1

6=2 t+3

2

51) x2

x2−4= xx+2

− 2x2−x 52)

3x+26

−5−2 x12

=2+3 x3

53) x

x2+3 x+4+ x+1x2+6x+8

= 2 xx2+x−2

54) 12x2=2 x

3

55) 1+ x−1x−3

= 2x−3

−x 56) x−4x−3

= x−52 x−6

57) 2c15

−23=1

3− c

5 58) x−9

2x−5+ 7−8 x

10−4 x=5

2

59) 16c2−4

= c−2c+2

+ c+22−c 60)

3x−2

= 4x2−2 x

+ 3x

61) a3=1a−1

6 62) 2

x2−5x+6− 3

3+2 x−x2 =1

x2−x−2

285

63) 1

(c+11) (2c+3 )= 1

8 c3+2764)

x−2x−7

= x+32x−14

65) x2+x−12x3−3 x2−2 x+3

= 16 x−9 66)

4 x+1116x+12

−1−4 x8 x+6

= 4 x+34

67) 2 x+3

x2+2 x−3+ x−3

2 x2−3 x+1= 5x

2 x2+5 x+368)

2x−12

+ 2x+13 x

=32

69) x+5

2x2−3 x−2+

( x−1 )( x2−5x+6 )

= 3 x+42 x2−5 x−3 70)

2x−16

−5 x+112

=12

71) x3=5

6− 1

2 x 72) 3 x2−x=4

73) x5+ 3x=8

5 74) 3 x2=4 x

75) 4

3t+12− 5

8−2 t= 2 t+3t 2−16

76) 3x5

−1=2 x3

+2

77) p+1

18 p+12= 2

9 p2−4+ 1

4−6 p 78) 2x3

+4=3 x2

−23

References


Boston: Pearson.






Company.

286

Answers to previous problems on the following pages.

287

Solutions to “Solve by Factoring” Problems

1) x=−12

∨x=7 2) x=0∨x=32∨x=2

3) x=−73

∨x=0 4) u=−4

3∨u=0∨u=2

5) x=−1∨x=6 6) n=−12

∨n=0∨n=32

7) x=−2∨x=5 8) m=−2∨m=0∨m=34

9) x=−13

∨x=0∨x=12 10) x=−3∨x=1

3

11) x=2∨x=3 12) x=32∨x=2

13 x=−5∨x=−2 14) x=0∨x=3

15) x=−3∨x=2 16) x=−52

∨x=23

17) t=−2∨t=8 18) x=−13

∨x=2

19) m=32∨m=2 20) x=0must reject x=3

21) x=−3∨x=−23 22) x=

−14

∨x=3

23) g=−4∨g=3 24) x=−2∨x=34

25) p=1∨p=2 26) x=−12

27) x=−2∨x=32 28) x=

67

29) w=−13

∨w=3 30) t=247

31) x=−34

∨x=2 32) x=353

33) n=−75

∨n=32 34) x=−1

35) x=13∨x=3

2 36) x=−5∨x=−1

37) x=12∨x=5

2 38) x=−6∨x=6

39) x=−3∨x=3 40) x=5

41) x=−43

∨x=43 42) n≠

23

43) m=−6∨m=3 44) b≠13

45) t=−2∨t=9 46) c=−1247) a=−6∨a=−1 48) a=−3

288

49) x=4 50) t=−5

51) x=−1∨x=0 52) x=−94

53) x=−16 54) x=0∨x=4

355) x=−2 56) no solution

57) c=3 58) x≠52

59) no solution 60) no solution

61) a=−2∨a=32 62) x=

14

63) c=−14

∨c=2 64) no solution

65) x=−2∨x=12 66) x≠−3

4

67) x=43 68) x=

13∨x=1

69) x=83 70) x=−9

71) x=1∨x=32 72) x=−1∨x=4

3

73) x=3∨x=5 74) x=0∨x=43

75) t=−1011 76) x=−45

77) p=−4 78)x=285

References


Boston: Pearson.






Company.

289

Additional Solving Quadratic or Rational Problems with Detailed Solutions

Example: 2c15

−23=1

3− c

5

The most important strategy for rational expressions that can be performed is to

“clear fractions” by multiplying each term by the LCD (least common denominator). In

one swift motion, so to speak, all denominators are cancelled leaving only the

numerators. Proceed as follows:


1) 2c15

−23=1

3− c

5 Recopy equation, LCD is 15.

2) 15∗( 2c15

−23)=15(¿ 1

3− c

5) Multiply both sides of equation by

LCD.

3) 15∗2c

15−15∗2

3=15∗1

3−15∗c

5 Distribute “15” to each term.

4) 15∗2c

15−5∗3∗2

3=5∗3∗1

3−5∗3∗c

5 Decompose each term.

At this point, to see more clearly the steps involved in reducing fractions, make it

a point to use a different colored pen or pencil and cross out the terms that will remove

a factor of 1. For this application, the red colored numbers will remove a factor of 1,

leaving only the numerators in the expression.

5) 15∗2c

15−5∗3∗2

3=5∗3∗1

3−5∗3∗c

5

6) 2c−5∗2=5∗1−3∗c "Clearing fractions".

7) 2c−10=5−3c Intermediate result.

291

8) 2c−10+3c=5−3c+3c Add3c to each side.

9) 5c−10=5

10) 5c−10+10=5+10 Add 10 to each side.

11) 5c=15

12) 5c5

=155 Divide each side by 5.

13) c=3 Final answer

The most important step was to multiply each term by the LCD, which “cleared

out” fractions efficiently. This critical step will be performed for rational expressions in

order to make further computations and manipulations easier.

292

Example:x+5

2x2−3 x−2+ x−1x2−5 x+6

= 3 x+42x2−5 x−3


denominators at this time. The denominators are monic and non-monic trinomials,

which were discussed from previous sections. Proceed as follows:


1) x+5

(2x+1 ) ( x−2 )+ x−1

( x−2 ) ( x−3 )= 3 x+4

(2x+1 ) ( x−3 ) Factor the denominators.

Note LCD is (2 x+1)(x−2)(x−3)

At this point, make it a point to realize that to maintain balance in the entire

equation; the LCD must be multiplied to ALL terms!

2) (2 x+1 ) ( x−2 ) ( x−3 )∗[ x+5(2 x+1 ) ( x−2 )

+ x−1( x−2 ) (x−3 )

= 3 x+4(2 x+1 ) (x−3 )

] Multiply entire

expression by LCD.

3) (2 x+1 ) ( x−2 ) ( x−3 )∗x+5(2 x+1 ) ( x−2 )

+(2 x+1 ) ( x−2 ) (x−3 )∗x−1( x−2 ) ( x−3 )

=(2 x+1 ) ( x−2 ) ( x−3 )∗3x+4(2x+1 ) ( x−3 )

To see more clearly the steps involved in reducing fractions, make it a point to

use a different colored pen or pencil and cross out the terms that will remove a factor of

1. For this application, the red colored expressions will remove a factor of 1, leaving

only the numerators in the expression.

4)(2 x+1 ) ( x−2 ) ( x−3 )∗x+5(2 x+1 ) ( x−2 )

+(2 x+1 ) ( x−2 ) (x−3 )∗x−1( x−2 ) ( x−3 )

=(2 x+1 ) ( x−2 ) ( x−3 )∗3x+4(2x+1 ) ( x−3 )

293

Notice the denominators "cleared out", so to speak, leaving only the numerators.

This step is very critical for any rational expression in order to make further

computations easier.

5) ( x−3 ) ( x+5 )+(2 x+1 ) ( x−1 )=( x−2 )(3 x+4) Intermediate result.

At this time, FOIL needs to be performed on both sides of equation in order to

remove grouping symbols, which are the parenthesis.

6) x2+5x−3x−15+2 x2−2 x+ x−1=3 x2+4 x−6 x−8

7) 3 x2+x−16=3 x2−2 x−8 Like terms collected.

At this time, recognize the presence of a quadratic equation due to 3 x2 term. This

equation must be set equal to zero before further computations can be performed.

Proceed as follows:

8) 3 x2−3 x2+x−16=3 x2−3 x2−2x−8 Subtract3 x2 from each side of

equation.

9) x−16=−2 x−8 Intermediate result, which leaves

only linear equation. The3 x2

term has been cancelled.

At this time, some discussion is required to explain how to solve any linear

equation. The overall strategy is to separate terms containing variables from constants.

294

In other words, variables go to left side of equals sign, numbers with no variables go to

right side of equals sign.

10) x+2x−16=−2 x+2x−8 Add 2 x to each side of equation.

11) 3 x−16=−8 Intermediate result.

12) 3 x−16+16=−8+16 Add 16to each side of equation.

13) 3 x=8

14) 3x3

=83 Divide each side by3.

15) x=83 Final answer

295

Example:p+1

18 p+12= 2

9 p2−4+ 1

4−6 p


denominators at this time. Determining the Greatest Common Factor (GCF) for each

term’s denominator is required at this time. Proceed as follows:


1) p+1

6 (3 p+2 )= 2

(3 p−2 ) (3 p+2 )+ 1

2 (2−3 p ) Factor denominators.

Notice the denominator in the very last term has “factors that are opposite”.

Manipulate the last term’s denominator only in order to make further computations

easier.

2) p+1

6 (3 p+2 )= 2

(3 p−2 ) (3 p+2 )+ 1

2∗(−1)(−2+3 p ) Factor out a negative 1 from only

the denominator in last

term, paying close attention to

signs.

3) p+1

6 (3 p+2 )= 2

(3 p−2 ) (3 p+2 )+ 1−2 (3 p−2 ) Only working with last term,

multiplying−1∗2=−2, and

rewriting (−2+3 p ) as

(3 p−2).

4) p+1

6 (3 p+2 )= 2

(3 p−2 ) (3 p+2 )− 1

2 (3 p−2 ) Again, only working with last term

in order to take care of minus

sign located in

296

denominator. Commit to

memory “a negative

divided by a positive equals a

negative.”

Now, determine the LCD for entire expression, which follows to be

6 (3 p+2 ) (3 p−2 ) . In order to “clear out” fractions, multiply each term by LCD.

5) 6 (3 p+2 ) (3 p−2 )∗[ p+16 (3 p+2 )

= 2(3 p−2 ) (3 p+2 )

− 12 (3 p−2 ) ] Multiply entire

expression

by LCD.

6)6 (3 p+2 ) (3 p−2 )∗( p+1 )6 (3 p+2 )

=6 (3 p+2 ) (3 p−2 )∗2(3 p−2 ) (3 p+2 )

−6 (3 p+2 ) (3 p−2 )∗12 (3 p−2 )

Notice how the LCD is multiplied by each term. This step is critical for any type of

rational expression. At this point, to see more clearly the steps involved in reducing

fractions, make it a point to use a different colored pen or pencil and cross out the terms

that will remove a factor of 1. For this application, the red colored expressions will


7)6 (3 p+2 ) (3 p−2 )∗( p+1 )6 (3 p+2 )

=6 (3 p+2 ) (3 p−2 )∗2(3 p−2 ) (3 p+2 )

−3∗2 (3 p+2 ) (3 p−2 )∗12 (3 p−2 )

8) (3 p−2 )∗( p+1 )=6∗2−3∗(3 p+2)

9) (3 p−2 ) ( p+1 )=12−3(3 p+2)

297

10) 3 p2+3 p−2 p−2=12−9 p−6 FOIL on left side of equation,

distributing on right side. Small

steps are necessary in order to

understand the process.

11) 3 p2+p−2=6−9 p Collect like terms.

At this point, it is critical for any quadratic equation like the one above; to be set

equal to zero before any further manipulations can be performed.

A quadratic equation of the form a x2+bx+cis present, therefore, collect all terms

to one side of equals sign. This is referred to as “setting equation equal to zero”.

Proceed as follows:

12) 3 p2+p−2=6−9 p Rewrite equation for clarity.

13) 3 p2+p+9 p−2=6−9 p+9 p Add 9 p to each side of equation.

14) 3 p2+10 p−2=6 Collect like terms.

15) 3 p2+10 p−2−6=6−6 Subtract 6 from both sides.

16) 3 p2+10 p−8=0

Notice how the equation is now “set equal to zero”. At this point, a non-monic

trinomial is present, and factoring this type is as follows:

17) 3 p2+10 p−8=0 Recopy problem, noting mentally the

10p term

Decompose first term, do same for last

term.

298

3p

p

4

2

8

1





practice, the process becomes extremely efficient. Proceed as follows:

First product :3 p∗4=12 p

Second product : p∗2=2 p

∑ :12 p+2 p=14 p

Difference :12 p−2 p=10 p

At this time, the sum or difference of the two products does indeed equal 10p.

The two products are12 p∧2 p. This non-monic trinomial is factorable. Proceed as

follows:

18) ()()=0 "Open, close, open, close".

19) (3 p ) (p )=0 Insert 3 p∧p from first column.

20) (3 p ) (p 4 )=0 Remember the pair 3 p∧4 are mated

together to represent the OUTER


299

4

2

3p

p

21) (3 p2 ) ( p4 )=0 Remember the pair p∧2 are mated

together to represent the INNER portion

of the FOIL process.

22) (3 p2 ) ( p4 )=0

At this time, the signs need to be inserted correctly. The question that arises is

"Where do you put the signs and why?" Remember12 p−2 p=10 p, therefore, the

positive sign will be placed in front of the4 , the negative sign will be placed in front of

the2. This will ensure a positive10 p.

23) (3 p−2 ) ( p+4 )=0 Signs inserted correctly.

At this time, set each factor equal to zero. This property is called the zero factor

property.



24) (3 p−2 ) ( p+4 )=0

25) 3 p−2=0∨p+4=0

Work with each factor individually at this time. Proceed as follows:

26) 3 p−2+2=0+2 p+4−4=0−4

27) 3 p=2 p=−4

28) 3 p3

=23

29) p=23

300

30) p=23∨p=−4

At this time, it is critical that both answers be checked to be sure no extraneous

solutions exist. For this problem, p=23 is extraneous due to the fact it would cause

denominator to equal zero, therefore p=23 cannot be used as one of the answers, or in

other words, rejectp=23 . The final answer isp=−4.

301

Example:x−4x−3

= x−52 x−6

Rational expressions are present, therefore, factor denominators if possible, and

then determine the LCD.


1) x−4x−3

= x−52( x−3) Factor

LCD is2 ( x−3 )

2) 2(x−3)∗x−4

x−3=2 ( x−3 )∗x−5

2(x−3)Multiply each term by LCD.






3) 2(x−3)∗x−4

x−3=2 ( x−3 )∗x−5

2(x−3)

4) 2 ( x−4 )=x−5 Cleared fractions.

5) 2 x−8=x−5 Distribute. a (b+c )=ab+bc

6) 2 x−x−8=x−x−5 Subtract x from both sides of equation.

7) x−8=−5 Intermediate result.

8) x−8+8=−5+8 Add 8 to both sides of equation.

9) x=3

302

At this point, realize that 3 cannot be used as a viable answer due to the fact it

would cause denominator to equal zero. Therefore, no solutions exist.

10) no solution Final answer

303

Example:2c+1c+3

=2c

Rational expressions are present, therefore, factor denominators if possible, and

then determine the LCD.

LCD:c (c+3)


1) 2c+1c+3

=2c Recopy equation.

2)c( c+3 )∗2c+1

c+3=c (c+3 )∗2

cMultiply each term by LCD.






3) c( c+3 )∗2 c+1

(c+3)=c (c+3 )∗2

c

4) c (2c+1 )=2(c+3) Clear fractions.

5) 2c2+c=2 c+6 Distribute. a (b+c )=ab+bc

6) 2c2+c−2 c=2c−2c+6 Subtract 2c from both sides.

7) 2c2−c=6 Intermediate result.

8) 2c2−c−6=6−6 Subtract 6 from both sides.

9) 2c2−c−6=0

304

At this time, the quadratic equation is now "set equal to zero", and further

computations can be performed. In addition, a non-monic trinomial is present and can

be factored as follows:

Decompose first and last term into factors.





practice, the process becomes extremely efficient. Proceed as follows:

First product :2c∗2=4 c

Second product :3∗c=3 c

∑ : 4c+3c=7 c

Difference :4 c−3c=c

305

2c

c

6

1

2

3

2c

c

2

3

At this time, the sum or difference of the two products does indeed equal the

middle term. The two products are4 c∧3c. This non-monic trinomial is factorable.

Proceed as follows:

10) ()()=0 “Open, close, open, close”.

11) (2c ) ( c )=0 Insert2c∧cfrom first column.

12) (2c ) ( c2 )=0 Remember 2c∧2are mated

together to represent the OUTER

portion of FOIL.

13) (2c3 ) (c 2 )=0 Remember 3∧care mated

together to represent the INNER portion

of FOIL.

At this time, the signs need to be inserted correctly. The question that arises is

"Where do you put the signs and why?" Remember−4c+3c=−c, therefore, the positive

sign will be placed in front of the3and the negative sign will be placed in front of the4.

This will ensure a positive−c.



Set each factor equal to zero at this time.

14) (2c+3 ) (c−2 )=0

15) 2c+3=0∨c−2=0

16) 2c+3−3=0−3 c−2+2=0+2

17) 2c=−3 c=2

18) 2c2

=−32

306

19) c=−32

20) c=−32

∨c=2 Final answer

References

Bittinger, E. (2006). Elementary and Intermediate Algebra,Concepts and Applications.

Boston: Pearson.

Hall, E. (2010). College Algebra. Corpus Christi, TX: Hall, E.

Hall, E. (2010). Elementary Algebra. Corpus Christi, TX: Hall, E.

Hall, E. (2010). Intermediate Algebra. Corpus Christi, TX: Hall, E.



Company.

End of section

307

Appendix A: Complete Lesson Plans

The following lesson plans developed for this project are not available in print

form, but are included in the enclosed compact disc.

Lesson Plan 1: Divisibility Rules..........................................................................20

Step-by-Step Procedures....................................................................................22

Lesson Plan 2: Greatest Common Factor...........................................................30


Lesson Plan 3: Factoring by Grouping Method...................................................37


Lesson Plan 4: Sum of Perfect Squares..............................................................48


Lesson Plan 5: Difference of Perfect Squares.....................................................56


Lesson Plan 6: Sum of Cubes.............................................................................65


Lesson Plan 7: Difference of Cubes....................................................................73


Lesson Plan 8: Monic Trinomials.........................................................................81


Lesson Plan 9: Non-Monic Trinomials.................................................................92


Lesson Plan 10: Techniques of Factoring.........................................................107

Step-by-Step Procedures..................................................................................109

308

Lesson Plan 11: Solving an Equation for a Certain Variable.............................111

Step-by-Step Procedures..................................................................................113

309

Appendix B: Student Packet

The following student packet developed for this project is not available in print

form, but is included in the enclosed compact disc.

Formulas............................................................................................................134

Factoring tips.....................................................................................................136

Table of perfect squares and perfect cubes......................................................137

List of prime numbers........................................................................................138

Rules of divisibility.............................................................................................140

Polynomial flowchart..........................................................................................141

Divisibility rule worksheets.................................................................................142

Greatest common factor problems (5)...............................................................144

Factoring by grouping problems (5)...................................................................145

Difference of square problems (5).....................................................................146

Sum of square problems (5)..............................................................................147

Sum of cube problems (5).................................................................................148

Difference of cube problems (5)........................................................................149

Monic trinomial problems (5).............................................................................150

Non-monic trinomial problems (5)......................................................................151

Solve by factoring problems (5).........................................................................152

Putting it all together problems (5).....................................................................153

310

Appendix C: Student Packet Answer Key

The following student packet answer key developed for this project is not

available in print form, but is included in the enclosed compact disc.

Greatest common factor answers (5)................................................................156

Factoring by grouping method answers (5).......................................................156

Difference of perfect square answers (5)...........................................................156

Sum of square answers (5)...............................................................................157

Sum of cube answers (5)...................................................................................157

Difference of cube answers (5)..........................................................................157

Monic trinomial answers (5)...............................................................................158

Non-monic trinomial answers (5).......................................................................158

Solve by factoring answers (5)..........................................................................158

Putting it all together answers (5)......................................................................159

311

Appendix D: Hyperlinks for Videos

The following hyperlinks for the video tutorials developed for this project are

included in the enclosed compact disc and are directly linked to the community

college’s website.

Hyperlinks for videos.........................................................................................161

312

Appendix E: Additional Problems and Solutions Available for Students

The following additional problems and solutions developed for this project are

not available in print form, but are included in the enclosed compact disc.

Greatest common factor problems, answers, and detailed solutions................163

Factoring by grouping method problems, answers, and detailed solutions.......180

Difference of perfect square problems, answers, and detailed solutions...........193

Sum of square problems, answers, and detailed solutions...............................206

Sum of cube problems, answers, and detailed solutions...................................212

Difference of cube problems, answers, and detailed solutions..........................225

Monic trinomial problems, answers, and detailed solutions...............................241

Non-monic trinomial problems, answers, and detailed solutions.......................257

Solve by factoring problems, answers, and detailed solutions..........................281

313

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