8

8
Factoring Polynomials
Part 2
Learner’s Module in Mathematics 8
Quarter 1 Week 2

Cecile T. Duran
Developer
Department of Education • Republic of the Philippines

NAME: _____________________________ GRADE & SECTION: ______________

TEACHER: __________________________ SCORE: ________________________
 Republic of the Philippines
DEPARTMENT OF EDUCATION
Cordillera Administrative Region
SCHOOLS DIVISION OF BAGUIO CITY
Military Cut off, Baguio City

Published by:
DepEd Schools Division of Baguio City
Curriculum Implementation Division

COPYRIGHT NOTICE
2020

Section 9 of Presidential Decree No. 49 provides:

“No copyright shall subsist in any work of the Government of the Philippines.
However, prior approval of the government agency of office wherein the work is
created shall be necessary for exploitation of such work for profit.”

This material has been developed for the implementation of the K-12 Curriculum
through the DepEd Schools Division of Baguio City – Curriculum Implementation
Division (CID). It can be reproduced for educational purposes and the source must
be acknowledged. Derivatives of the work including creating an edited version, an
enhancement or a supplementary work are permitted provided all original work is
acknowledged and the copyright is attributed. No work may be derived from this
material for commercial purposes and profit.

ii
 PREFACE

This module is a project of the DepEd Schools Division of Baguio City through
the Curriculum Implementation Division (CID) which is in response to the
implementation of the K-12 Curriculum.

This Learning Material is a property of the Department of Education, Schools

Division of Baguio City. It aims to improve students’ academic performance
specifically in Mathematics.

Date of Development : August 2020

Resource Location : DepEd Schools Division of Baguio City
Learning Area : Mathematics
Grade Level :8
Learning Resource Type : Module
Language : English
Quarter/Week : Q1/W2
Learning Competency/Code : (1) Factors completely different types of
polynomials (perfect square trinomials, quadratic
trinomials and grouping)
(M8AL-Ia-b-1)

: (2) Solves problems involving factors of

polynomials (M8AL-Ib-2)

iii
 ACKNOWLEDGEMENT

The developer wishes to express her gratitude to those who helped in the
development of this learning material. The fulfillment of this learning material would
not be possible without them.
Thank you all for your moral and technical support in the crafting of this
learning module. To my colleagues in our school for sharing your knowledge and
expertise as I developed this learning resource. Lastly, to the office of DepEd
Division of Baguio City for giving us the opportunity to discover our skills as module
writers.

Development Team
Author/s: Cecile T. Duran
Layout Artist / Editors: Melchor B. Ticag and Aiza R. Bitanga
Illustrator:

School Learning Resources Management Committee

Brenda M. Cariño School Principal
Editha L. Laop Subject/ Learning Area Specialist
Niño E. Martinez Subject/ Learning Area Specialist
Sherwin Fernando School LR Coordinator

Quality Assurance Team

Francisco C. Copsiyan EPS – Mathematics
Niño M. Tibangay PSDS – BCNHS District

Learning Resource Management Section Staff

Loida C. Mangangey EPS – LRMDS
Victor A. Fernandez Education Program Specialist II – LRMDS
Christopher David G. Oliva Project Development Officer II – LRMDS
Priscilla A. Dis-iw Librarian II
Lily B. Mabalot Librarian I
Ariel Botacion Admin. Assistant

CONSULTANTS

JULIET C. SANNAD, EdD

Chief Education Supervisor – CID

SORAYA T. FACULO, PhD

Asst. Schools Division Superintendent

MARIE CAROLYN B. VERANO, CESO V

Schools Division Superintendent

iv
 TABLE OF CONTENTS

COPYRIGHT NOTICE..................................................................................................ii
PREFACE.....................................................................................................................iii
ACKNOWLEDGEMENT...............................................................................................iv
TABLE OF CONTENTS................................................................................................v
Learning Objectives......................................................................................................1
Pre-assessment............................................................................................................1
Introduction...................................................................................................................3
Lesson4.......................................................................................................................4
Perfect Square Trinomial...........................................................................................4
Exercise 1: Factor the given polynomials..............................................................5
Lesson5.......................................................................................................................6
Factoring Quadratic Trinomials................................................................................6
Exercise 2: Factor the given polynomials............................................................10
Lesson6.....................................................................................................................10
Factoring by Grouping.............................................................................................10
Exercise 3: Factor the given polynomials............................................................11
Generalization ……………………………………………………………………………..12
Application..................................................................................................................13
Activity 1: Find the Hidden Message...................................................................13
Post-assessment........................................................................................................13
Performance Task......................................................................................................15
Activity 2: Solving word problems involving factoring..........................................15
ANSWER KEY............................................................................................................16
REFERENCES...........................................................................................................17

v
 Learning Objectives
Hello learner! This module was designed and written with you in mind.
Primarily, its scope is to develop your understanding in factoring different types of
polynomials: perfect square trinomials, quadratic trinomials, and factoring by
grouping.

While going through this module, you are expected to:

1. determine the types of factoring used in the given;
2. factor the given polynomials completely using perfect square trinomial,
quadratic trinomial, and grouping; and
3. solve problems involving factoring polynomials.

Note
This module will serve as your personal copy. You will write your FINAL
ANSWERS on the ANSWER SHEET attached with this module. You will only
submit the answer sheet. However, I suggest that you still write your answers
on the spaces provided for your answers, so you can check your answers once
you receive the answer key. You may also use the margins for your solutions /
computations.

Pre-assessment
This test will just determine how much you already know about the lessons. I
suggest that you just answer what you know and don’t worry about your score.
It is very important that you answer the test before you proceed to the next part of
this module. DO NOT attempt to answer any unanswered items nor change your
answers in this test after you have learned about the lessons in this module. Please
be honest.

DIRECTION: Read and understand each item, then choose the letter of the correct
answer and write it on your answer sheet.

1. One binomial factor of m2−11 m+ 21is m−3. What is the other factor?
A. m−7 C. m+8
B. m+7 D. m−8

2. What are the factors of s ( s−3 )+ 8 ( s−3 ) ?

A. ( s−3 ) and ( s−8 ) C. ( s−3 ) and ( s+8 )
B. ( s+3 ) and ( s+8 ) D. ( s+3 ) and ( s−8 )

3. What will you multiply to 5 x+ 1to make a perfect square trinomial?

A. 5 x−1 C. 25 x 2+10 x +1
B. 5 x+ 1 D. 25 x 2−10 x+1
4. Which pair of numbers, when multiplied together, yields the product 4
2
p +8 p+3?
A. ( 2 p−1 ) and ( 2 p−3 ) C. ( 2 p−1 ) and ( 2 p+ 3 )
B. ( 2 p+ 1 ) and ( 2 p+ 3 ) D. ( 2 p+ 1 ) and ( 2 p−3 )

5. What is the missing term in _____−80 n+64 to form a perfect square

trinomial?
A. 25 n2 C. 30 n2
2
B. 20 n D. 35 n2

6. If one factor of 8 k 2 ( k −2 )−( k−2 ) is the binomial k −2 , what is the other factor?
A. 8 k 2 C. 8 k 2 +1
2
B. 8 k −0 D. 8 k 2 −1

7. From what factors can we get the product s2 +4 s−21?

A. ( s−7 ) and ( s−3 ) C. ( s+7 ) and ( s+3 )
B. ( s−7 ) and ( s+3 ) D. ( s+7 ) and ( s−3 )

8. What is the length of the side of a square whose area is

( 4 x 2+20 x +25) cm 2?
A. ( 2 x−5 ) cm C. ( 2 x+25 ) cm
B. ( 2 x+5 ) cm D. ( 2 x−25 ) cm

9. What are the factors of 3 b ( b−4 )−( b−4 ) ?

A. ( b−4 ) ( 3 b+1 ) C. ( b−4 )2 (3 b−1 )
B. ( b−4 ) ( 3 b−1 ) D. 3 b ( b−4 )2

10. Which factors yields to 5 c−ct+5 d−dt ?

A. ( c−d )( 5−t ) C. ( c +d ) ( 5−t )
B. ( c−d )( 5+t ) D. ( c +d ) ( 5+t )

11. Which of the polynomial expression can be factored by perfect square

trinomial?
A. x 2−5 x+ 6 C. x 2+ 10 x +25
B. x 2−10 x−25 D. x 2−14 x−20

12. One of the factors of 2 p 2−13 p +15 is p−5 . What is the other factor?
A. 2 p−3 C. 2 p+ 3
B. p−3 D. p+5

13. Which of the options represents the factored form of x 2+ 5 x +6 ?

A. ( x +2 ) ( x +3 ) C. ( x +4 ) ( x +2 )
B. ( x +5 ) ( x+1 ) D. ( x +3 ) ( x+3 )

14. Which from the choices is the factored form of 49 j 2−56 j+16 ?
A. ( 7 j−4 ) (7 j+ 4 ) C. ( 7 j+ 4 ) ( 7 j+ 4 )
2
B. ( 7 j+ 4 ) D. ( 7 j −4 )2

15. What are the factors of a 2+ ab−2 a−2 b ?

 A. ( a−2 ) ( a+b ) C. ( a+ 2 )( a+ b )
B. ( a−2 ) ( a−b ) D. ( a+ 2 )( a−b )
Introduction
Let’s recall how much you have learned in the previous lessons on common
monomial factoring, factoring difference of two squares, and factoring sum and
difference of two cubes by answering the activity that follows.

Review: Complete Me

Factoring is the __________ of multiplication.

Direction: To complete the statement above, match each polynomial expression in

the table to its factored form found in the answer box. Write the corresponding letter
of your answer on the space provided for.

1. 2. 3. 4.

3 a+3 b+ 3 c 2
a −b
2 3
a −64 10 ab−2 ac

5. 6. 7.

2
b 3+125 2
a b+3 a b
2 2 a −16

Answer Box

R ( b+5 ) ( b2−5 b+25 )E ( a+ b ) ( a−b ) S a 2 b ( 1+3 b ) E 2 a ( 5 b−c )

R 3 ( a+ b+c ) V ( a−4 ) ( a2 + 4 a+16 ) E ( a+ 4 )( a−4 )

In this module, you will continue learning how to factor polynomial

expressions completely. The first three types / methods of factoring polynomial
expressions were already discussed in the previous week (common monomial
factoring, factoring difference of two squares, and factoring sum and difference of
two cubes). For this week, you will tackle the last three methods (factoring perfect
square trinomials, factoring quadratic trinomials, and factoring by grouping).

Lesson
4 Perfect Square Trinomial

A polynomial is a perfect square trinomial if it satisfies these conditions:

1. the polynomial is a trinomial (three terms);
2. the first and the last terms are perfect squares, and are positive; and
3. the middle term is twice the product of the square roots of the perfect
square terms and may either be positive or negative.

To factor a Perfect Square Trinomial (PST), get the square root of the first term,
copy the sign of the middle term, and get the square root of the last term. The
Squared Binomial is the factored form of the Perfect Square Trinomial (PST).

RULE or
or

Example 1. Factor x 2+ 6 x+ 9.

Solution: Step 1: Determine if the polynomial is a perfect square trinomial.

2
x + 6 x+ 9 The first term ( x ¿¿ 2)¿ and the last term( 9 ) are
perfect squares.
x3 √ x 2=x since x • x=x 2
√ 9=3 since 3 •3=9
2 ( x )( 3 ) =6 x
The middle term ( 6 x ) is twice the product of the
square roots of the first and the last terms.

Step 2: Substitute in the pattern x 2+ 2 xy + y 2= ( x + y )2

since the middle term is positive.

x 2+ 6 x+ 9=( x +3 )2∨(x+ 3)(x +3)

square root of
sign of the middle term
square root of

2
Checking: ¿ ( x +3 ) Apply the rule in squaring a binomial.
2 2
Square the first term, get twice the product of the
¿ x + 2 ( x ) ( 3 ) +3 first and the last terms, then square the last term.
2
¿ x + 6 x+ 9 Simplify.
Conclusion: Since the obtained product in the checking is equal to the original
polynomial expression, therefore, x 2+ 6 x+ 9=( x +3 )2.

Example 2. Factor 49 x 2−14 x +1.

Solution: Step 1: Determine if the polynomial is a perfect square trinomial.

49 x 2−14 x +1 The first term (49 x 2 ) and the last term (1) are
perfect squares.
7 x1 √ 49 x 2=7 x since 7 x • 7 x =49 x 2
√ 1=1 since 1 •1=1
2 ( 7 x ) ( 1 )=14 x
The middle term (14 x ) is twice the product of
the square roots of the first and the last terms.
Step 2: Substitute in the pattern x 2−2 xy+ y 2=( x− y )2
since the middle term is negative.
2 2
49 x −14 x +1= (7 x−1 ) ∨(7 x−1)(7 x−1)
square root of
sign of the middle term
square root of

Checking: ¿ ( 7 x−1 )2 Apply the rule in squaring a binomial.

Square the first term, get twice the product of
2 2 the first and the last terms, then square the last
¿ ( 7 x ) + 2 ( 7 x ) (−1 )+ (−1 )
term.
2 Simplify.
¿ 49 x −14 x +1

Conclusion: Since the obtained product in the checking is equal to the original
polynomial expression, therefore, 4 9 x2 −14 x+ 1=( 7 x−1 )2.

Exercise 1: Factor the given polynomials.

1) c 2−12 c +36

2) 64 y 2 +80 y +25

3) 25 r 2−30 r + 9

4) 9 a 2+12 a+ 4

5) 36 w 2−12 w+1
 Lesson
5 Factoring Quadratic Trinomials

There are some trinomials that are not the result of a square of a binomial,
thus, we cannot use the pattern for perfect square trinomial.
This time, we are going to factor a quadratic trinomial in the form

Ax2 + Bx+C .

Factoring Trinomials of the form , where

5.a
RULE
where and

Find two factors of whose sum is equal to .

Example 1. Factor x 2+ 5 x + 4.

Solution: Step 1. Determine the two factors of 4 whose sum is equal to 5.

Factors of 4 Product Sum

2 and2 2(2)=4 2+2=4
4 and 1 4 (1)=4 4 +1=5

Note: The table will just serve as a guide in finding the factors of C
whose sum is B in the trinomial of the form A x 2+ Bx+ C . You may
or may not do the same as long as you can find immediately the
needed factors. You will notice in the succeeding examples that
there are some numbers with many possible factors to consider.

Step 2. Substitute in the pattern A x 2+ Bx+ C=( x+ v ) ( x + w ) .

x 2+ 5 x + 4= ( x+ 4 ) ( x +1 ) Carry the signs of + 4 and +1.

Checking: ¿ ( x +4 ) ( x +1 ) Multiply the obtained factors.

¿ ( x • x ) + ( x •1 ) + ( 4 • x )+(4 •1) Apply the FOIL method.
2
¿ x + x+ 4 x+ 4 Combine similar terms.
 2
¿ x +5x+4

Conclusion: Since the obtained product in the checking is equal to the original
2
polynomial expression, therefore, x + 5 x + 4= ( x+ 4 ) ( x +1 ).
Example 2. Factor m2−7 m+10.

Solution: Step 1. Determine the two factors of 10 whose sum is equal to −7 .

Factors of 10 Product Sum
10 and 1 . 10(1)=10 10+1=11
2 and5 2(5)=10 2+5=7
. −10 and −1 (−10)(−1)=10 .−10+(−1)=−11

−2 and−5 (−2)(−5)=10 −2+(−5)=−7

Step 2. Substitute in the pattern A x 2+ Bx+ C=( x+ v ) ( x + w ) .

2
m −7 m+10= ( m−2 ) ( m−5 ) Carry the signs of −2 and −5.

Checking: ¿ ( m−2 ) ( m−5 ) Multiply the obtained factors.

¿ ( m• m )+ ( m•−5 ) + (−2• m) +(−2 •−5) Apply the FOIL method.

2
¿ m −5 m−2 m+ 10 Combine similar terms.
2
¿ m −7 m+10

Conclusion: Since the obtained product in the checking is equal to the original
2
polynomial expression, therefore, m −7 m+10= ( m−2 ) ( m−5 ) .

Example 3. Factor s2 +11 s−12.

Solution: Step 1. Determine the two factors of −12 whose sum is equal to 11.
Factors of −12 Product Sum
−6 and 2 (−6)(2)=−12 −6+ 2=−4
−4 and 3 (−4)(3)=−12 −4+3=−1
−12and1 (−12)(1)=−12 . −12+1=−11
6 and −2 (6)(−2)=−12 6+(−2)=4
4 and −3 4 (−3)=−12 4 +(−3)=1
12and−1 ( 12 ) (−1 )=−12 12+(−1)=11
 Step 2. Substitute in the pattern, A x 2+ Bx+ C=( x+ v ) ( x + w ) .
2
s +11 s−12=( s+12 )( s−1 ) Carry the signs of +12 and −1.

Checking:¿ ( s+12 ) ( s−1 ) Multiply the obtained factors.

¿ ( s • s ) + ( s •−1 ) + ( 12• s ) +(12 •−1) Apply the FOIL method.
2
¿ s −s+12 s−12 Combine similar terms.
2
¿ s +11 s−12

Conclusion: Since the obtained product in the checking is equal to the original
2
polynomial expression, therefore, s +11 s−12=( s+12 )( s−1 ) .

Example 4. Factorw 2−2 w−24 .

Solution: Step 1. Determine the two factors of −24 whose sum is equal to −2.

Factors of −24 Product Sum

−1and 24 (−1)(24)=−24 −1+24=23
−2and 12 (−2)(12)=−24 −2+12=10
−3and 8 (−3)(8)=−24 −3+8=5
−4and6 (−4)(6)=−24 −4+6=2
1and −24 (1)(−24)=−24 1+(−24)=−23
2and −12 (2)(−12)=−24 2+(−12)=−10
3and −8 (3)(−8)=−24 3+(−8)=−5
4 and−6 ( 4)(−6)=−24 4 +(−6)=−2

Step 2. Substitute in the pattern, A x 2+ Bx+ C=( x+ v ) ( x + w ) .

2
w −2 w−24=( w+ 4 )( w−6 ) Carry the signs of + 4 and −6 .

Checking:¿ ( w +4 )( w−6 ) Multiply the obtained factors.

¿ ( w • w ) + ( w •−6 ) + ( 4 •w ) +(4 •−6) Apply the FOIL method.

2
¿ w −6 w+ 4 w−24 Combine the middle terms.
2
¿ w −2 w−24
Conclusion: Since the obtained product in the checking is equal to the original
polynomial expression, therefore, w 2−2 w−24=( w+ 4 )( w−6 ) .

There are quadratic trinomials in the form Ax2 + Bx+C where A is not equal to
one . In this case, the above method in factoring x 2+ Bx +C cannot be applied. Let’s
study the next rule when A is greater than one ( A>1).
 Factoring Trinomials of the form , where

where and
5.b
RULE Factor the first term, and the last term, .
Look for two factors such that the sum of their product is the middle term,

Note that we still apply the concept of factoring where is positive and
negative.

Example 5. Factor 3 x 2+7 x + 4.

Solution: Step 1. Factor the first and the last terms.

Terms Factors
3x
2
3 x, x
4 2 , 2 and 4 , 1
Step 2. Determine the factors of 3 x 2and 4 whose sum is equal to the middle
term, 7 x .
Possible Factors Middle Term
( 3 x+ 2 )( x +2 ) 8x
( 3 x+ 4 )( x +1 ) 7x
( 3 x+ 1 )( x +4 ) 13 x
Checking: ¿ ( 3 x+ 4 )( x +1 )
Multiply the obtained factors.
¿ ( 3 x • x ) + ( 3 x • 1 )+ ( 4 • x ) + ( 4 •1 ) Apply the FOIL method.
2
¿ 3 x +3 x +4 x+ 4 Combine similar terms.
2
¿ 3 x +7 x + 4
Conclusion: Since the obtained product in the checking is equal to the original
polynomial expression, therefore, 3 x 2+7 x + 4= ( 3 x + 4 ) ( x+1 ) .

Example 6. Factor 9 n2 −5 n−4.

Solution: Step 1. Factor the first and the last terms.

Terms Factors
9 n2 3 n , 3 n and 9 n , n
−4 −2 , 2 ; 4 ,−1 and −4 , 1
 Step 2. Determine the factors of 9 n2 and −4 whose sum is equal to the
middle term, −5 n.
Possible Middle Middle
Possible Factors
Factors Term Term
( 3 n−2 )( 3 n+2 ) 0 ( 9 n+2 )( n−2 ) −16 n
( 3 n+ 4 ) ( 3 n−1 ) 9n ( 9 n+ 4 )( n−1 ) −5 n
( 3 n−4 )( 3 n+1 ) −9 n ( 9 n−4 ) ( n+1 ) 5n
( 9 n−2 ) ( n+2 ) 16 n

Checking:¿ ( 9 n+ 4 ) ( n−1 ) Multiply the obtained factors.

¿ ( 9 n •n )+ ( 9 n•−1 )+ ( 4 • n ) + ( 4 •−1 )Apply the FOIL method.
2
¿ 9 n −9 n+4 n−4 Combine similar terms.
2
¿ 9 n −5 n−4
Conclusion: Since the obtained product in the checking is equal to the original
polynomial expression, therefore,9 n2 −5 n−4=( 9 n+4 ) ( n−1 ) .
Exercise 2: Factor the given polynomials.

1. b 2+8 b +15

2. f 2−11 f +18

3. p2−3 p−18

4. z 2+ 8 z−9

5. 5 y 2+ 29 y−42

Lesson
6 Factoring by Grouping

Let’s recall that the product of two binomials having no common terms is
composed of four terms and that most of the time, we use the Distributive Property of
Multiplication over Addition.

Steps in Factoring by Grouping:

1. Group the terms that have common factors. Apply ¿ ( ax +ay ) + ( bx +by )
the commutative and associative properties of
addition.
2. Factor out the GCF of each grouped terms. ¿ a ( x + y ) +b ( x + y )
3. Factor out the common binomial factor. Write it as ¿ ( x + y ) (¿¿)
 the first factor.
4. Get the algebraic sum of the GCF of each grouped ¿ ( x + y )( a+b )
terms. Write it as the second factor.

To factor a polynomial by grouping, we follow this pattern

RULE
.

Example 1: Factor 2 x−10+ xy−5 y .

Solution: ¿ ( 2 x−10 ) + ( xy−5 y ) Group the terms that have common factors.
¿ 2 ( x−5 ) + y ( x−5 ) Factor out the GCF of each grouped terms.
¿ ( x−5 ) ¿ Factor out the common binomial factor.
¿ ( x−5 ) ( 2+ y ) Divide each term by the common binomial.
Checking:¿ ( x−5 ) ( 2+ y ) Multiply the obtained factors.
¿ ( x • 2 )+ ( x • y ) + (−5 •2 )+ (−5• y ) Apply the FOIL method.

¿ 2 x+ xy −10−5 y Rearrange the terms.

¿ 2 x−10+ xy−5 y
Conclusion: Since the obtained product in the checking is equal to the original
polynomial expression, therefore, 2 x−10+ xy−5 y= ( x−5 ) ( 2+ y ).

Example 2: Factor 6 c +3 ac−10−5 a.

Solution: ¿ ( 6 c +3 ac )−( 10+5 a ) Group the terms that have common factors.
¿ 3 c ( 2+ a )−5 ( 2+ a ) Factor out the GCF of each grouped terms.
¿ ( 2+a ) ¿ Factor out the common binomial factor.
¿ ( 2+a )( 3 c−5 ) Divide each term by the common binomial.

Checking:¿ ( 2+a )( 3 c−5 ) Multiply the obtained factors.

¿ ( 2 •3 c ) + ( 2•−5 )+ ( a •3 c ) + ( a •−5 ) Apply the FOIL method.

¿ 6 c−10+ 3 ac−5 a Rearrange the terms.

¿ 6 c +3 ac−10−5 a
Conclusion: Since the obtained product in the checking is equal to the original
polynomial expression, therefore, 6 c +3 ac−10−5 a=( 2+ a ) ( 3 c−5 ) .

Example 3: Factor ( a+ b ) ( a−3 ) + ( a+b )( a−4 ) .

Solution: ¿ ( a+ b ) ¿ Factor out the common binomial factor.

¿ ( a+ b ) [ ( a−3 )+ ( a−4 ) ] Divide each term by the common binomial.
¿ ( a+ b ) ( 2 a−7 ) Combine similar terms.

Thus,( a+ b ) ( a−3 ) + ( a+b )( a−4 ) =( a+b )( 2 a−7 ) .

Exercise 3: Factor the given polynomials.

1. ax +ay +2 x+ 2 y

2. 4 s +4 t +5 rs +5 rt

3. 3 c−cp+3 d−dp

4. g ( g+1 )−7 ( g+1 )

5. ( d +2 ) ( d−3 )+ ( d+ 2 )( d+2 )
 Generalization

To summarize the entire lessons on Factoring Polynomial Expressions, both

modules 1 and 2, let us consider the table below.

Type of Factoring Rule

Common Monomial Factoring ax +ay + az=a ( x + y + z )

Factoring Difference of Two Squares 2 2

x − y =( x + y ) ( x− y )

x + y =( x + y ) ( x −xy + y )
3 3 2 2
Factoring Sum or Difference of Two Cubes
x − y = ( x − y ) ( x + xy + y )
3 3 2 2

2 2 2
Factoring Perfect Square Trinomial x + 2 xy + y = ( x + y )
2 2 2
x −2 xy+ y =( x− y )

2
Factoring Quadratic Trinomial x + Bx +C=( x + v)(x +w)
2
Ax + Bx+C=( px+ q ) ( rx + s )

Factoring by Grouping ax +ay +bx +by =( a+b )( x + y )