Module 17
The XYZ Factors
What this module is all about
This module is about factoring polynomials. This deals on products of polynomials
whose terms have a common monomial factor, trinomials which are products of two
binomials, trinomials which are squares of a binomial, and products of the sum and
difference of two quantities. From time to time, you will be referred to the lessons in the
preceding module on special product because special product and factoring are reverse
processes. You will see also how factoring can be applied widely in many mathematics
problems.
This module includes the following lessons:
Lesson 1
Lesson 2
Lesson 3
Lesson 4
Lesson 5
Lesson 6
Factoring and Common Factors
Factoring Expressions of the Form x2 + bx + c
Factoring Expressions of the Form ax2 + bx +c
Factoring Perfect Square Trinomial
Factoring the Difference Between Two Squares
Sum and Difference of Two Cubes
What you are expected to learn
After working on this module you should able to:

Factor polynomials in each of the following cases:





polynomials whose terms have a common monomial factor
trinomials which are products of binomials
trinomials which are squares of a binomial
difference of two squares
sum and the difference between two cubes
1
How to learn from this module
This is your guide for the proper use of the module:
1. Read the items in the module carefully.
2. Follow the directions as you read the materials.
3. Answer all the questions that you encounter. As you go through the module, you
will find help to answer these questions. Sometimes, the answers are found at
the end of the module for immediate feedback.
4. To be successful in undertaking this module, you must be patient and
industrious in doing the suggested tasks.
5. Take your time to study and learn. Happy learning!
Start
Take the Pretest
Check your paper and count
your correct answers.
Is your score
80% or above?
Yes
Scan the items you
missed.
No
Study this module
Take the Posttest
Proceed to the next
module/STOP.
2
What to do before (Pretest)
I. Multiple Choice. Choose the letter of the correct answer.
1. If one of the factors of 18x5y + 12x4y – 6x3y + 3x2y is 3x2y, what is the other
factor?
a. 6x4 + 4x3 – 2x2 + x
b. 6x3 + 4x2 – 2x + 1
c. 6 + 4x – 2x2 + x3
d. 6x + 4x2 – 2x3 + x4
2. Which of the following expressions has (x – 4) and (x + 3) as factors?
a. x2 + 7x + 12
b. x2 + x – 12
c. x2 – 7x – 12
d. x2 – x – 12
3. Which of the following is TRUE?
a.
b.
c.
d.
25x2 – 120xy + 144y2 = (5x – 12y)2
9a2 – 24a – 16 = (3a – 4)2
16x2 + 12xy + 9y2 = (4x + 3y)2
x2 – y2 = (x – y)2
4. Which of the following is a factor of 9a2 – 25b2?
i.
3a – 5b
a. i
b. ii
ii. 3a + 5b
c. i and ii
d. cannot be determined
5. Which of the following are factors of a3 – 8b3?
a.
b.
c.
d.
a – 2b; a2 – 4ab + b2
a – 2b; a2 – 2ab + 4b2
a – 2b; a2 + 4ab + 4b2
a – 2b; a2 + 2ab + 4b2
II. Determine the factors of the following polynomials.
A.
1. 3x+ 12
2. 10m + 6m2
3
3. xy2z3 – x2y3z4
4. ay + by – cy + 3y
5. x2 – 49
6.
1 2 9 2
a  b
4
16
7. x4 – y4
8. (a – b)2 – 16
B.
1. a2 + 14a + 49
2. m2 + 6n + 8
3. t2 – 15t + 16
4. x6 – 27
5. 8a3 + 125
6.
1 6
r 8
27
7. 2x2 – 5x – 12
8. 4n2 – 15n + 9
C.
Answer these questions:
a. If the area of a square is 25x2 + 30x + 9, what is the measure of its side?
b. A rectangle has an area of x2 + 12x + 32. If the length is represented by x +8, what is
the width of the rectangle?
Answer Key on page
What you will do
Lesson I
Factoring and Common Factor
In the previous module, you studied about special ways of finding the product
of polynomials. The results are called special products. This time, we will do the other
way around. You will be given the product and you will determine its factors. This process
4
is factoring. It is the reverse process of multiplying. When we factor expressions, we
determine an equivalent expression that is a product of two or more expressions.
Before we proceed further, let us review what you have learned about
factors, prime factorization and greatest common factor.
Following are some examples that will enable you to recall some concepts
you have studied before.
1. Factor: a) 36
b) 54
c) 72
Solution: For each given number, we may write the factorization process in a
form of factor tree as shown here:
36
9
x
54
4
9 x
3 x 3x2 x 2
72
6
9
3 x 3X 2 x 3
x
8
3 x 3x 2 x 2X2
Look at the darkened numbers in the last row. They are the prime
factorization of 36, 54 and 72.
Why do you think we call these the prime factorization of the given number?
We say that if a given number is written as a product of prime factors, they
are factored completely.
From the prime factorization as shown , we can determine the Greatest
Common Factor (GCF) of the numbers 36, 54, and 72.
36
54
72
GCF
= 3
= 3
= 3
= 3
= 18
x
x
x
x
3
3
3
3
x
x
x
x
2
2
2
2
x 2
x 3
x 2
x 2
We say that 18 is the largest number that could divide 36, 54 and 72 exactly.
It is also called their greatest common divisor.
The concept of GCF can now be extended to monomials and polynomials.
5
Study the following examples:
1. Find the GCF of 4x2, 8x4 and 12x6
SOLUTION:
We factor each of the given monomials where the highest common
factor can be determined among the three monomials.
4x2 = 4x2  1
8x4 = 4x2  2x2
12x 6 = 4x2  3x4
GCF = 4x2
The greatest common factor of 4x2, 8x4, and 12x6 is 4x2.
2. Find the GCF of 8m3n4 – 20m2n3 -10mn2
8m3n4
=
2
3
-20m n =
-10mn2 =
GCF
=
2mn2 x 4m2n2
2mn2 x -10mn
2mn2 x - 5
2mn2
The processes of prime factorization and finding the GCF will be very useful
in solving problems involving factoring.
Remember:
Factoring is the process of obtaining equivalent expression that
is the product of two or more expressions.
A. Common Monomial Factor
10x2
Suppose you are given a rectangle whose area is represented by
+15xy, with a width of 5x, how do we express the length of this rectangle?
A situation like this can be solved with the knowledge of common monomial
factor.
Study the illustrative examples below on factoring polynomials whose
terms have a common monomial factor.
6
Recall that in multiplying
monomials and polynomials; you use the
distributive property. That is, you multiply the monomial to each of the terms in
the polynomial factor.
To illustrate:
x ( b + c +d) = bx +cx + dx.
As such, x is a common factor of bx +
cx + dx.
This time, we do the reverse. We look into the polynomials product
bx + cx + dx and determine the factor common among the terms.
Examples:
1. Factor 9y4 – 15y3 + 3y2
Solution:
 Find the monomial of highest degree in the polynomial which can be factored
from each term.
Consider both the literal factors and the numerical factors.
a. The lowest degree of the literal factors in the terms 9y4, -15y3 and 3y2 is y2
and therefore y2 is
the highest power which can be factored from the
literal factors in each term.
b. Find the GCF of the numerical coefficient 9, 15 , 3
GCF (9, 15, 3 ) = 3
c. Find the product of the greatest numerical factor and literal factor.
Therefore, the greatest common monomial factor is 3y2.
9y4
3y2  3y2

–
15y3
3y2  –5y
+
3y2
3y2 1
Divide each term by 3y2 to get the terms of the remaining polynomial factor.
3y2 ( 3y – 5y + 1)
Therefore: 9y4 – 15y3 + 3y2 = 3y2 ( 3y2 – 5y + 1 )
7
2.
Factor : 6a + 8a2b – 14a3b2
= (2a  3) + (2a  4ab) – (2a  7a2b2)
(Factoring out the greatest common
factor 2a)
= 2a (3 + 4ab – 7a2b2)
(Dividing each term by the GCF 2a to
get the remaining polynomial factor )
3.
Factor: 18x5y + 12x4y – 6x3 + 3x2y
Solution:
18x5y + 12x4y – 6x3 + 3x2y
=(3x2y  6x3) + (3x2y  4x2) – (3x2y  2x) + (3x2y  1)
( Factoring out 3x2y)
= 3x2y (6x3 + 4x2 – 2x + 1)
( Dividing each term by
3x2)
Self-check 1
I. Factor the following expressions.
1. 15x2 + 21y2
2. 12x2 + 20xy
3. 15xy2 + 25xy
4. 13a2b – 26ab2
5. 14abc + 28a3b3c3
II. Answer the following problem:
1. If the area of a rectangle is represented by 6a2 + 15ab, and its width is 3a,
what is the length of this rectangle?
2. The perimeter of the rectangle below is 10a + 6b and the width as
indicated . How do you represent the measure of the length of the rectangle?
a+b
Answer Key on page 34
8
Lesson 2 Factoring Expressions of the Form x2 + bx + c
In Lesson 1, you were asked to factor polynomials whose terms have common
monomial factor. Here, you will be factoring trinomials of the form
ax2 + bx
+c
where a = 1.
If a = 1, the trinomial will also be in the form x2 + bx + c.
We may begin with a review of finding the product of binomial, say :
( x+6)(x+2) = ?
Do you still remember the FOIL Method? It will be useful when factoring polynomials
in the form x2 + bx + c as shown here:
F
O
I
L
(x + 6) (x + 2) = x2 + 2x + 6x + 12
= x2
+
8x
+
12
If we do the reverse process, that is, given the product, how do we determine the
factors?
In the polynomial product, x2 + 8x + 12 , you cannot find any common
factor
among the 3 terms. Therefore, factoring as discussed in Lesson 1 does not apply here.
Instead, another way of factoring this trinomial is applied.
To factor x2 + 8x + 12, think of the reverse of the FOIL Method. Referring to
the
above example, note the following:

x2 is the result of x  x. Therefore, the first term of each binomial factor is x.
We can write (x + _ ) (x + _)

The coefficient of the middle term of the trinomial are two numbers whose
product is 12 and whose sum is 8.
Here, you first think of two numbers when you multiply will give 12.
We may write these pairs of numbers in a table.
9
Product of 12
Sum
4, 3
7
-4, -3
-7
6, 2
8
-6, -2
-8
1, 12
13
-1, 12
-13
The right column contains the sum of the pairs of factors of 12. Ask yourself
“What are the factors of +12 that will give the sum of 8?”
The 3rd row satisfies the
two conditions : product is 12 and sum is 8.
The numbers we need are 6 and 2.
We now write the factorization:
x2 + 8x + 12 = (x +6) (x + 2)
Example 2:
Factor y2 – 9 y + 20
Since the last term is positive and the middle term is negative, it means we
need two negative numbers whose product is 20. and whose sum is -9.
Product of 20
Sum
-1. -20
-21
1, 20
21
,5
9
-4, -5
-9
2, 10
20
-2, -10
-20
The numbers are -4 and -5.
So, y2 -9y + 20 = (y-4) (y-5)
10
We may now check :
(y – 4) (y – 5) = y2 – 5y – 4y +20
= y2 – 9y +20
In the preceding examples, C , the constant term is positive or c> 0.
However, there are cases when the constant term of the trinomial is negative
or C < 0.
Example 3: Factor x2 + 3x -10
In this case, we look for two numbers whose product is the last term, -10 and
whose sum is equal to the numerical coefficient of the middle term, +3 . Since the
constant term is -10, it has a positive factor and a negative factor. The mental
process maybe written this way:
Thus,
Product -10
Sum
-1, 10
9
1, -10
-9
2, -5
-3
-2, 5
3
x2 + 3x – 10 = (x-2)(x+ 5)
We check:
( x – 2 ) ( x +5 ) = x2 + 5x -2x -10
= x2 + 3x - 10
Example 4 : Factor: y2 -17y – 18
Find two numbers whose sum is -17 and whose product is -18
11
Product -18
9, -2
-9, 2
-1, 18
1, -18
-3, 6
3, -6
Sum
7
-7
17
-17
3
-3
Thus, y2 -17y – 18 = ( y + 1)(y -18)
Self-check 2
I. Factor each of the following:
1. a2 + 8a + 15
2. p2 – 3p – 10
3. x2 – x – 20
4. y2 + 7y – 18
5. b2 – 5b + 6
II. Answer these problems:
1. A balloon flies at a speed of ( x – 2) miles per hour . At this rate, how long
will it take to fly ( x2 + 6x – 16) miles?
2. If the area of a rectangle is m2 + 4m +3, what two binomials could
represent the length and the width of the rectangle?
Answer Key on page 35
Lesson 3 Factoring Expressions of the Form ax2 + bx +c
In Lesson 2, you were introduced to factoring trinomials of the form
ax2 + bx + c where a = 1.
In the following examples, you will see how to factor trinomial of the form
12
x2 + bx + c where
used here.
a ≠ 1 . The skills you have developed in the previous lesson will be
Consider this product:
( 2x + 1 ) ( 3x + 2 )
F
0
I
L
= 6x2 + 4x + 3x + 2
= 6x2 + 7x
Factors
+2
As you can see, the numerical coefficient of the first term of the two
binomial
factors is not equal to 1. As such, the numerical coefficient of the first
term
of
the
trinomial product is not equal to 1 also. This leads to more combinations
of possible
factors.
Like in Lesson 2, think of the reverse of the FOIL Method to determine the factors
of a given polynomial. To illustrate how these trinomials are factored, here are some
examples:
Example 1. Factor 5x2 – 18x – 8
Solution:
Step 1. Determine if there is a factor common to all terms. There is none.
Step 2. Look at the first terms. Determine two numbers whose product is 5.
1, 5
–1, –5
Step 3. Now, look for numbers whose products is -8.
–4, 2
1, –8

;
;
4, –2
8, –1
Since the last term is negative, the second terms in each factor must be of
different signs. Here are some possible combinations in the factorization.
a.
b.
c.
d.
e.
f.
g.
h.
( x + 2) ( 5x – 4)
( x + 1) ( 5x –8)
( 5x – 1) ( x + 8)
( 5x + 1) ( x – 8)
( x – 1) ( 5x + 4)
( x – 1) ( 5x + 8 )
( x – 4) ( 5x + 2 )
( x + 4) ( 5x – 2)
13
In each case, when we multiply the first terms, the product is 5x2, and the last erm is –8.
Step 4. Look again at the 8 possible combinations. If you apply the distributive
property
and combine like terms, which pair will give us the middle term
–18?
The combination in d gives us the correct middle term. Thus.
5x2 – 18x – 8 = ( x – 4) (5x + 2)

Check:
( x – 4) (5x + 2) = 5x2 + 2x – 20x – 8
= 5x2 – 18x – 8
Example 2: Factor 2x2 – 7x – 4
Think :
For the first terms: Find two numbers whose product is 2.
For the last terms: Find two numbers whose product is –4
Determine possible factorizations among the combination of factors:
a.
b.
c.
d.
e.
( 2x – 4) ( x + 1)
( 2x – 1) ( x + 4)
( 2x + 1) ( x – 4)
( 2x + 2) ( x – 2)
( 2x – 2) ( x + 2)
From the above combination, select the pair which will satisfy this condition:


The outside product plus the inside product must equal –7x.
Combination c satisfies the given condition.
Example 3: Factor 20 + 6x – 2x2
20 + 6x – 2x2 = 2 ( 10 + 3x – x2)
Factoring the common factor 2
Yes, you have already factored. But the product as written may not be completely
factored yet. Examine the polynomial factor:
10 + 3x – x2
14
.
This polynomial can still be factored!
Think:
First terms: Find two numbers whose product is 10
Last terms: Find two numbers whose product is –x2
Determine the possible factorization:
a.
b.
c.
d.
( 5 + x) ( 2 – x)
( 5 – x) ( 2 + x)
(10 + x) ( 1 – x)
(10 – x) ( 1 + x)
The outside product plus the middle product must equal 3x.
Combination b satisfies this condition.
Thus, the complete factorization is:
20 + 6x – 2x2 = 2 ( 5 – x) ( 2 + x)
Self-check 3
A. The questions below are given to help you factor 4x2 – x – 18. Answer each
question.
1. The product of the numerical coefficient of the first term (4x2) and the last term
(-18) is -72. What are the possible pairs of factors of -72?
2. The factors -9 and 8 give a sum of -1.
How do you write 4x2 – x – 18 such that -x is expressed in terms of -9 and 8?
Supply the missing part in the statement below.
3. 4x2 4.
– 18 = (4x2 – 9x) + (8x – 18)
+
(4x – 9) +
5. (4x – 9) (
+
( 4x – 9)
)
15
B. Factor the following expressions.
1. 2a2 + a – 15
2. 3x2 – 8x – 3
3. 2x2 – 5x - 12
4. 3x2 + x – 10
5. 12x2 – 5x – 2
C. Answer this problem:
1. The area of the rectangle below is 6x2 + 22x + 20 and the other measures are
shown in the figure. What are the measures of a and b?
3x
5
Lesson 4 Factoring Perfect Square Trinomial
Let’s recall the process of squaring a binomial.
(x
=
+
4 )2
x2 + 2(4)(x) + 42
x2 + 8x + 16
(x
–
6 )2
x2 – 2(6)(x) + 62
16
= x2 – 12x + 36
(5x
+
4 )2
(5x)2 + 2 (5x)(4) + 42
=25x2 + 40x + 16
In the above examples, we have squared the binomials ( x + 4), ( x – 6), and
(5x + 4). When you square a binomial the resulting product is a perfect square
trinomial. It is also called a trinomial square. The two factors of perfect square
trinomial are exactly alike.
How do we check if a given trinomial is a perfect square?
can
A perfect square trinomial can be recognized by inspection and the factors
be identified and written at once.
Following are perfect square trinomials:
1.
2.
3.
4.
5.
x2 + 14x + 49
4y2 – 12y + 9
25a2 + 20a + 4
9n4 + 30n2 + 25 = ( 3n + 5)2
64 – 16x + x2 = ( 8 – x )2
Examine again the above trinomials.



What can you say about the first and the third term in each trinomial?
What is the sign of the first and third term in each?
How is the middle term of the trinomial related to the factor of the first
and third term of the polynomial?
We can now write:
If a trinomial is perfect square, the first and last terms are perfect squares
and the middle term whether positive or negative is twice the product of the
square roots of the first and last terms. It may be represented in the form:
a2 ± 2ab + b2
17
The trinomials given above are perfect squares for the following reasons:
__ __
2
1. x + 14x + 49 because 14x = 2√x2 √ 49
= 2 (x)(7)
= 14x
2. 4y2 -12y +9
__ __
because -12y = 2√4y2 √9
= 2 (2y)(3)
= 12y
___ __
3. 25a2 +20a +4 because 20a = 2√25a2 √4
= 2(5a)(2)
= 20a
Apply this test to trinomial d and e.
Examine how factoring is done in each polynomial given below
1.
x2 + 14x +
49
x2
49
x
7
 Since middle term is +,
x2 + 14x + 49 = (x + 7)2
18
2.
4y2 –
12y
+
9
4y2
9
2y
3
 Since the middle term is –,
4y2 -12y + 9 = (2y – 3)2
3.
25a2
+
20a
+
4
25a2
4
5a
2
Since the middle term is +,
25a2 + 20a + 4 = (5a + 2)2
4. 9n4 + 30n2 + 25 = (3n + 5)2
5. 64 – 16x + x2 = (8 – x )2
Check the factoring done in numbers 2, 3, 4 & 5 by finding the special product
(square of a binomial).
Remember:
To factor a perfect square trinomial, get the square root of the first and last
term. The sign of the middle term of the trinomial determines the sign of the
binomial factor. Since factors are identical, express the square of the binomial
factor.
19
Self-check 4
A. Tell whether each of the following is a perfect square trinomial or NOT. Give
reason for your answer.
1. x2 + 12x + 36
3. 9a2 – 12ab + 4b2
2. x2 + xy + y2
4 . 4x2 + 10xy + 25y2
2
5. x – x + 1
B. Factor the following polynomials:
1. a2 + 6a + 9
4. 25z2 + 10z + 1
2
2. b – 16b + 64
5. x2 – 3x – 4
3. 4y2 – 4y + 1
C. Answer the following problems:
Refer to the figure below:
1. If the area of a square ABCD is 16a2 + 40a + 25 units, what is the measure of
its side? What is the perimeter of this square?
2. Given the side a+3, what is the area of the smaller square?
3. What is the total area of the two squares?
4. What is the total distance around the two squares?
A
B
E
16a2 + 40 a + 25
D
a+3
C
F
G
20
Lesson 5 Factoring the Difference Between Two Squares
Here, you will be introduced to another type of factoring. This is the easiest type of
factoring. As mentioned at the beginning of this module, factoring is associated with finding
the
product,
that
is,
the
reverse
process
of
multiplying.
21
Recall finding the special product in this form:
( x + 8) ( x – 8 )
= x2 – 64
( 2x + 4) ( 2x – 4) = 4x2 – 16
( 3a + 5) ( 3a – 5) = 9a2 – 25
The product of the binomial on its left can be described as the difference
between two squares. You see that the first and the second term are perfect
squares and the sign between them is minus.
Likewise, the above examples illustrate the pattern for factoring polynomial
expressed as difference between two squares.
Here are other examples:
x2
1.
–
100
x2
100
x
10
 ( x + 10) ( x – 10)
2.
a2 –
144
a2
144
a
12
 (a + 12 ) (a – 12 )
3. 9m2 – 49
= ( 3m + 7) ( 3m – 7)
4. 16x4 – 225
= ( 4x2 + 15) ( 4x2 – 15)
22
5. x8 – 16
= ( x4 + 4) ( x4 – 4 )
= ( x4 + 4) ( x2 + 2) ( x2 – 2)
(Factoring further x4 - 4 )
We now generalize how to factor an expression written as difference of two
squares:
(First term)2 – (Second term)2 =(First term + Second term) (First term – Second term)
Factoring difference of two squares lends itself to other numerical computations.
Study the examples below:
68 x 72
( 70 – 2) x ( 70 + 2)
= 702 – 4
Apply special product on difference between two squares
= 4900 – 4
Multiply
= 4896
Subtract
Find the product of 68 and 72 by the long method.
Give your own example. Check the results.
Self-check 5
I. Factor each of the following
1. a2 – 16
2. 9x2 – 4
3. 64c2 – 1
4. 100y2 – 49z2
5. y2 – 81
II. Simplify by applying special product and factoring.
23
Compare the results.
a. 49 x 51
b. 17 x 23
c. 99 x 101
d. 44 x 56
III. Solve:
1. A 5m x 5m landscaping is to be done in one corner of a 20m x 20m field.
Find the area of the field that was not affected by the project.
5m
5m
20m
20m
Lesson 6
Sum and Difference of Two Cubes
The expressions a3 + b3 and a3 - b3 represent polynomials that are sum of two
cubes and difference of two cubes, respectively.
How do we express a3 + b3 as a product of two factors?
It can be done this way:
Let us divide a3 + b3 by a + b by the usual way of dividing a polynomial by a
polynomial:
24
 b3
3
b
a2
- ab
3
ab a 
3
2
a  ab
2
- ab
2
- ab ab
2
ab b
2
3
2
3
ab b
0
To check, we multiply the quotient and the divisor:
( a + b ) ( a2 – ab + b2) = a3 – a2b + ab2 + a2b – ab2 + b3
= a3 – a2b + ab2 + a2b – ab2 + b3
= a 3 + b3
The product gives the dividend in the problem!
Therefore:
(a + b ) ( a2 – ab + b2) = a3 + b3
or
(a3 + b3) = (a + b ) ( a2 – ab + b2)
Similarly, if we divide a3 – b3 by (a – b ), we get
a2 
3
ab a
3
 b2
3
b
ab
2
a - ab
2
 ab
2
2
 ab - ab
2
3
2
3
 ab b
 ab b
0
25
To check:
( a - b ) ( a2 + ab + b2) = a3 + a2b + ab2 - a2b – ab2 + b3
= a3 + a2b + ab2 - a2b – ab2 + b3
= a 3 - b3
Therefore:
(a - b ) ( a2 + ab + b2) = a3 + b3
or
3
3
(a - b ) = (a - b ) ( a2 + ab + b2)
Let us go back and summarize the two processes we have done:
a3 + b3 = ( a + b ) ( a2 – ab + b2)
a3 – b3 = ( a - b ) ( a2 + ab + b2)
Take note of the patterns of the factors:



In each case, one factor contains two terms, the other factor contains three
terms.
The first binomial factor is made up of the cube roots of the two cubes in the
binomial expression. The sign between them is the same as the sign between
the cubes.
The second factor, the trinomial, in each case is made up of the squares of
the cube roots, with the product of the cube roots as the middle term. Further
note that if it is sum of two cubes, the middle term of the trinomial factor is
negative ( – ). If it is the difference of the two cubes, the middle term of the
trinomial factor is positive ( + ). In either case, the two factors have only one
minus ( – ) sign.
To further illustrate these, examine the following examples:
Example 1: Factor : a3 + 8
[ sum of two cubes]
26
Solution:
a3
+
3
8
a3 +
a
3
+
8
2
opposite sign
(a
+
–
2 ) (a2
2a
+
square of the
first term of the
22 )
square of the second term
of the first factor
first factor
negative of product of the first term and
second term of the first factor
= (a + 2 ) ( a2 – 2a + 4)
Example 2: Factor 8x3 + 27y3
Solution:
8x3 + 27y3 = ( 2x)3 + (3y)3
=
[ 2x + 3y ] [ (2x)2 – (2x)( 3y) + (3y)2]
Opposite sign
=
( 2x + 3y) (4x2 – 6xy + 9y4)
Example 3: Factor 64x3 – 1
Solution:
64x3 – 1 = ( 4x – 1)[ (4x)2 + 4x( 1) + 12]
opposite sign
= (4x- 1) (16x2 + 4x + 1)
27
Self-check 6
I. Factor the following.
1.
2.
3.
4.
5.
1 + 8x3
y3 + 64
a3 – 125
125y3 – 8z3
27y3 – 125
Answer Key on page 36
Lesson 7
Factoring by Methods of Grouping and Combination of
Various Types
Some polynomials where terms have no common factor if taken as a
whole. A
common factor maybe present in two terms and the remaining two terms having another
common factor.
For example, ax + x + 4y + ay
You could see that if taken as 1 group, the four terms do not contain any common
factor. However, x is common to both first and second term, and y,
common to the
third and fourth term. In such a case, the polynomial can be separated into groups of terms
which have a common factor.
That is:
ax + 4x + 4y + ay = ( ax + 4x) + ( 4y + ay)
= a ( a+ 4) + y ( 4 + a)
= x( a + 4) + y (a + 4)
= (a + 4)(x + y)
Here are other examples:
28
Factoring each binomial
Commutative Property of Addition
for 4 + a
Factoring out the common factor
a+4
1. Factor : x3 – 2x2 – 3x + 6
Solution:
x3 – 2x2 – 3x + 6 = ( x3 – 2x2 ) – ( 3x – 6)
= x2 (x – 2) – 3(x – 2)
= (x
– 2) (x2 – 3)
Grouping of terms. Take note
of the change of sign from +6 to -6.
Why do you think so?
Factoring each binomial by finding
the common factor.
Factoring out x-2
So,
x2 – 2x2 – 3x + 6 = (x – 2) (x2 – 3)
2. Factor: 4m3 – 4m – 3m2+3
This example illustrates the combination of two types of factoring polynomial
by grouping of terms and by difference of two squares.
Solution:
4m3 -4m – 3m2 +3 = (4m3- 4m) – (3m2 -3)
= 4m(m2 -1) – 3(m2 -1)
Grouping of terms
Finding common factor of each
binomial
= (m2-1)(4m-3)
Factoring out m2 -1
Still factorable!
= (m+1)(m-1)(4m-3)
Factoring by difference of two
squares
Following are more examples of mixed cases. In each case, polynomial is
completely factored.
a. Factor : x3 + y3 + x + y
Solution:
x3 + y3 + x + y
= (x3 + y3) + ( x + y)
Grouping of terms
= ( x + y) ( x2 – 2xy + y2) + ( x + y)
Factoring by sum of two cubes
= ( x + y ) ( x2 – 2xy + y2 + 1)
Factoring out x + y
29
b. Factor: a3 + b3 + a2 – b2
Solution:
a3 + b 3 + a2 – b2
= ( a3 + b3) + (a2 – b2)
Grouping of terms
= ( a + b) ( a2 – ab + b2) + ( a + b ) ( a – b)
Factoring by sum of two
cubes and difference of
two squares
Factoring out a + b
= ( a + b) ( a 2 – ab + b2 + a – b)
c. Factor : x2 + 6x + 9 – y2
Solution:
x2 + 6x + 9 – y2
= ( x2 + 6x + 9) – y2
Grouping of terms
= ( x + 3 )2 – y2
=[(x+3)+y][(x+3)–y]
Rewrite the first terms as a
perfect square
Factoring as difference between
two squares
= ( x + 3 + y) ( x + 3 – y )
Simplifying
d. Factor: 10ax2 – 15ax – 25a.
Solution:
10ax2 – 15ax – 25a.
= 5a(2x2 – 3x –5)
Factor out common monomial
factor. 5a is common,
= 5a(2x2 – 5x + 2x – 5)
Factor
= 5a (2x2 – 5x) + (2x – 5)
Grouping terms
= 5a [x(2x – 5)] + (2x – 5)
= 5a (2x – 5) (x + 1)
30
e. Factor: -5b2 – 40bc – 80c2.
Solution:
= -5(b2 + 8bc + 16c2)
Factor out -5 since it is a
common monomial factor.
Factoring perfect square
trinomial
= -5(b + 4c)(b + 4c)
Self-check 7
I. Factor each of the following completely.
1. 3x2 – 3y2
2. 4a3 – 100a
3. 4y2 + 4y – 8
4. m – 64m3
5. n3 + n + n2 + 1
II. Solve:
1. The volume (lwh) of each rectangular prism below is expressed as a
polynomial. Find three factors to represent the length, width and height of the
prism.
x3 + 8x2 + 15x
31
What to do after (Posttest)
I. Multiple Choice. Choose the letter of the correct answer.
1. What is the greatest common factor of 20xy2, 16x2y3 and 36x3y4?
a. 4xy
c. 4x2y3
b. 4xy2
d. 4x2y2
2. Which of the following are factors of 2x2 + 7x + 6?
c. (2x + 3)(x – 2)
d. (2x – 3)(x + 2)
a. (2x + 3)(x + 2)
b. (2x – 3)(x – 2)
3. If one of the factors of 20x4 – x2y2 – 2x2y2 – 3xy3 – 4y4 is 5x – 4y, what is the other
factor?
a. 4x3 + 3x2y + 2xy2 + y3
b. 4x4 + 3x3y + 2x2y2 + y4
c. 4x2 + 3x + y + 1
d. 4x2 + 3x2y2 + 2xy + 1
4. Which of the following are the factors of x2 – 64y2?
a. (x – 8y)2
b. (x + 8y)(x + 8y)
c. (x – 8y)(x + 8y)
d. cannot be determined
5. Which of the following expressions has 7a + 2b and 2a + 4b as factors?
a. 14a2 + 32ab + 8b2
b. 14a2 + 28ab + 8b2
c. 14a2 + 24ab + 8b
d. 14a2 + 14ab + 8b
6. Which of the following is TRUE?
a.
b.
c.
d.
x4 – 4x2y2 + 2y4 = ( x2 – 2y2)2
a4 + 2a2b2 + 4a2 + b4 + 4b2 + 4 = ( a2 + b2 + 2)2
100x2 – 50xy + 25y2 = (10x – 5y)2
36x3 + 20xy + 49y2 = ( 6x + 7y)2
7. Which of the following is a factor of 8a3 – 27b3?
a.
b.
c.
d.
2a2 + 2ab + 3b2
4a2 + 6ab + 9b2
2a2 – 2ab + 3b2
4a2 + 6ab + 9b2
32
8. Which of the following is the correct factorization of 4x2 – 64y2?
a.
b.
c.
d.
4(x2 – 16y2)
(2x – 8y)(2x + 8y)
2(2x2 – 32y2
4(x – 4y)(x + 4y)
9. Find the factors of 9x2 + 12xy + 4y2
a.
b.
c.
d.
3x + 2y
3x – 2y
( 3x + 2y)2
( 3x – 2y)2
10. Which of the following are factor of b3 + ab2 – ab2 + 2a3 ?
c. b – a ; b2 – ab + a2
d. b – a ; b2 + ab + a2
a. b + a ; b2 + 2ab + 2a2
b. b + 2a ; b2 – 2ab + a2
II. Factor the following completely.
A.
1. 4x-12 + xy -3y
2. 8x2 + 10x – 3
3. -2x2 + x + 6
4. x2 – y2 +5x -5y
5. a2 – 4ab – 21b2
6. 8x3 – 125
7. 12x2 – x – 20
8. 2x2– 4x - 16
III. Solve the following problems:
1. If the area of a square is represented by the polynomial 64 – 122x + 49x2, how will
you represent the measure of the side of this square?
2. Drawn are two squares. Find the area of the shaded
region as shown at the right. Find the factors of this area.
2s
2s
3r
3r
33
Answer Key
Pretest
I.
1. b
2. d
3. a
4. c
5. d
II.
A.
1.
2.
3.
4.
5.
6.
3( x + 4)
2m ( 5 + 3m)
xyz ( yz2 – xy2z3
y ( a + b – c + 3)
( x + 7) ( x – 7)
(1 a + 3 b ) ( 1 a – 3 b)
2
4
2
4
7. ( x + y) ( x – y) ( x2 + y2)
8. (a – b + 4) ( a – b – 4)
B.
1.
2.
3.
4.
5.
6.
( a + 7)2
(m+ 4) ( m + 2)
not factorable
(x2 – 3) (x + 3x2 + 9)
( 2a + 5 ) ( 4a2 – 5a + 25)
(1 r2 – 2 ) ( 1 r4 + 2 r2 + 4 )
3
9
7. ( 2x + 3) ( x – 4)
8. ( 4n – 3) ( n – 3)
C.
a. S = 5x + 3
b. width = x + 4
Lesson 1 Self-Check
I. 1. 3(5x2 + 7y2)
2. 4x(3x + 5y)
3. 5xy(3y + 5)
II.
1. l = 2a + 5b
2. l = 6a + 2b
4. 13ab(a –2b)
5. 14abc(1 + 2a2b2c2)
34
Lesson 2 Self-Check
II.
1. a2 + 8a + 15 = (a + 3) (a + 5)
2. p2 – 3p – 10 = (p – 5) (p + 2)
1. (x + 8)
3. x2 – x – 20 = (x – 5)(x + 4)
2. (m + 1) (m + 3)
4. y2 + 7y – 18 =(y + 9)(y – 2)
5. b2 – 5b + 6 = (b – 2)(b – 3)
Lesson 3
A.
1. -1(72); 1(-72) ; 2(-36); -2(36); 3(-24); -3(24); 4(-18); -4(18);
6(-12); -6(12); 8(-9); -8(9)
2. 4x2 – 9x + 8x – 18
3. 4x2 – 9x + 8x – 18
4. x(4x – 9) + 2(4x – 9)
5. (4x – 9)( x + 2 )
B.
1. 2a2 + a – 15
= a + 3)(2a – 5)
4. 3x2 + x – 10
= (x + 2)(3x – 5)
2. 3x2 – 8x – 3
=(x – 3)(3x + 1)
5. 12x2 – 5x – 2
= (3x – 2)(4x + 1)
3. 2x2 – 5x – 12
= (x – 4)(2x + 3)
Lesson 4
A.
1.
2.
3.
4.
5.
perfect square trinomial
NOT
Perfect square trinomial
perfect square trinomial
NOT
B.
1.
2.
3.
4.
5.
C.
1. S= 4a + 5
P = 16a + 20
2. A =a2 + 6a + 9
3. 17a2 + 46a + 34
4. 17a + 16
35
(a + 3)(a + 3) or (a + 3)2
(b – 8)(b – 8) or (b – 8)2
(2y – 1)(2y – 1) or (2y –1)2
(5z + 1)(5z + 1) or (5z + 1)2
(x – 4)(x + 1)
Lesson 5
1. (a – 4)(a + 4)
2. (3x – 2)(3x + 2)
3. (8c – 1)(8c + 1)
4. (10y – 7z)(10y + 7z)
5. (y – 9)(y + 9)
Lesson 6
1. (1 + 2x)(1 – 2x + 4x2)
2. (y + 4)(y2 – 4y + 16)
3. (a – 5)(a2 + 5a + 25)
4. (5y – 2z)(25y2 + 10yz + 4z2)
5. (3y – 5)(9y2 + 15y + 25)
Lesson 7
1. 3x2 – 3y2
= 3(x – y)(x + y)
4. m – 64m3
= m(1 – 8m)(1 + 8m)
2. 4a3 – 100a
= 4a(a – 5)(a + 5)
5. (n3 + n) + (n2 + 1)
= (n2 + 1)(n + 1)
3. 4y2 + 4y – 8
= 4(y + 2)(y – 1)
Posttest
I.
1. b
2. a
3. a
4. b
5. a
6. b
7. a
8. b
9. c
10. b
II.
1. (x-3)(y+4)
2. ( 4x – 1 ) ( 2x + 3)
3. ( -x + 2 ) (2x + 3)
4. (x-y)(x+y+5)
5.
6.
7.
8.
III. 1. 8 - 7x units
2. 9r2- 4s2 = (3r – s)(3r+s)
END OF MODULE
36
(a – 7b) ( a + 3b)
(2x-5)(4x2+10x+25)
( 4x + 5) ( 3x – 4)
(2x-8)(x+2)