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Factoring Polynomials: Factoring Two Special Products

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

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Summary: This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. Factoring is an essential skill for success in algebra and higher level mathematics courses. Therefore, we have taken great care in developing the student's understanding of the factorization process. The technique is consistently illustrated by displaying an empty set of parentheses and describing the thought process used to discover the terms that are to be placed inside the parentheses. The factoring scheme for special products is presented with both verbal and symbolic descriptions, since not all students can interpret symbolic descriptions alone. Two techniques, the standard "trial and error" method, and the "collect and discard" method (a method similar to the "ac" method), are presented for factoring trinomials with leading coefficients different from 1. Objectives of this module: know the fundamental rules of factoring, be able to factor the difference of two squares and perfect square trinomials.

Note: Your browser may not currently support MathML. See our browser support page for additional details. You can always view the correct math in the PDF version.

Overview

  • The Difference of Two Squares
  • Fundamental Rules of Factoring
  • Perfect Square Trinomials

The Difference of Two Squares

Recall that when we multiplied together the two binomials ( a+b ) ( a b ) and ( a−b ) ( a b ) , we obtained the product a 2 − b 2 a 2 b 2 .

( a+b ) ( a−b )= a 2 − b 2 ( a b ) ( a b ) a 2 b 2

Perfect Square

Notice that the terms a 2 a 2 and b 2 b 2 in the product can be produced by squaring a a and b b, respectively. A term that is the square of another term is called a perfect square. Thus, both a 2 a 2 and b 2 b 2 are perfect squares. The minus sign between a 2 a 2 and b 2 b 2 means that we are taking the difference of the two squares.
Since we know that ( a+b ) ( a−b )= a 2 − b 2 ( a b ) ( a b ) a 2 b 2 , we need only turn the equation around to find the factorization form.

a 2 − b 2 =( a+b ) ( a−b ) a 2 b 2 ( a b ) ( a b )

The factorization form says that we can factor a 2 − b 2 a 2 b 2 , the difference of two squares, by finding the terms that produce the perfect squares and substituting these quantities into the factorization form.
When using real numbers (as we are), there is no factored form for the sum of two squares. That is, using real numbers,

a 2 + b 2 a 2 b 2 cannot be factored

Sample Set A

Example 1

Factor x 2 −16 x 2 16 . Both x 2 x 2 and 16 are perfect squares. The terms that, when squared, produce x 2 x 2 and 16 are x x and 4, respectively. Thus,

x 2 −16=( x+4 ) ( x−4 ) x 2 16 ( x 4 ) ( x 4 )

We can check our factorization simply by multiplying.

(x+4)(x-4) = x 2 -4x+4x-16 = x 2 -16. (x+4)(x-4) = x 2 -4x+4x-16 = x 2 -16.

Example 2

49 a 2 b 4 −121 49 a 2 b 4 121 . Both 49 a 2 b 4 49 a 2 b 4 and 121 are perfect squares. The terms that, when squared, produce 49 a 2 b 4 49 a 2 b 4 and 121 are 7 a b 2 7 a b 2 and 11, respectively. Substituting these terms into the factorization form we get

49 a 2 b 4 −121=(7 a b 2 +11) (7 a b 2 −11) 49 a 2 b 4 121 (7 a b 2 11) (7 a b 2 11)

We can check our factorization by multiplying.

(7a b 2 +11)(7a b 2 -11) = 49 a 2 b 4 -11a b 2 +11a b 2 -121 = 49 a 2 b 4 -121 (7a b 2 +11)(7a b 2 -11) = 49 a 2 b 4 -11a b 2 +11a b 2 -121 = 49 a 2 b 4 -121

Example 3

3 x 2 −27 3 x 2 27 . This doesn’t look like the difference of two squares since we don’t readily know the terms that produce 3 x 2 3 x 2 and 27. However, notice that 3 is common to both the terms. Factor out 3.

3 ( x 2 −9 ) 3 ( x 2 9 )

Now we see that x 2 −9 x 2 9 is the difference of two squares. Factoring the x 2 −9 x 2 9 we get

3 x 2 -27 = 3( x 2 -9) = 3(x+3)(x-3) 3 x 2 -27 = 3( x 2 -9) = 3(x+3)(x-3)

Be careful not to drop the factor 3.

Practice Set A

If possible, factor the following binomials completely.

Exercise 1

m 2 −25 m 2 25

Solution

( m+5 ) ( m−5 ) ( m 5 ) ( m 5 )

Exercise 2

36 p 2 −81 q 2 36 p 2 81 q 2

Solution

9 ( 2 p−3 q ) ( 2 p+3 q ) 9 ( 2 p 3 q ) ( 2 p 3 q )

Exercise 3

49 a 4 − b 2 c 2 49 a 4 − b 2 c 2

Solution

(7 a 2 +bc)(7 a 2 −bc) (7 a 2 +bc)(7 a 2 −bc)

Exercise 4

x 8 y 4 −100 w 12 x 8 y 4 −100 w 12

Solution

( x 4 y 2 +10 w 6 )( x 4 y 2 −10 w 6 ) ( x 4 y 2 +10 w 6 )( x 4 y 2 −10 w 6 )

Exercise 5

3 x 2 −75 3 x 2 −75

Solution

3 ( x+5 ) ( x−5 ) 3 ( x 5 ) ( x 5 )

Exercise 6

a 3 b 4 m−a m 3 n 2 a 3 b 4 m a m 3 n 2

Solution

am(a b 2 +mn)(a b 2 −mn) am(a b 2 +mn)(a b 2 −mn)

Fundamental Rules of Factoring

There are two fundamental rules that we follow when factoring:

Fundamental Rules of Factoring

  1. Factor out all common monomials first.
  2. Factor completely.

Sample Set B

Factor each binomial completely.

Example 4

4 a 8 b-36 b 5 . Factor out the common factor 4b. 4b( a 8 -9 b 4 ) 4 a 8 b-36 b 5 . Factor out the common factor 4b. 4b( a 8 -9 b 4 )

Now we can see a difference of two squares, whereas in the original polynomial we could not. We’ll complete our factorization by factoring the difference of two squares.

4 a 8 b-36 b 5 = 4b( a 8 -9 b 4 ) = 4b( a 4 +3 b 2 )( a 4 -3 b 2 ) 4 a 8 b-36 b 5 = 4b( a 8 -9 b 4 ) = 4b( a 4 +3 b 2 )( a 4 -3 b 2 )

Example 5

x 16 - y 8 . Factor this difference of two squares. x 16 - y 8 = ( x 8 + y 4 ) Sum of two squares Does not factor ( x 8 - y 4 )  Difference of two squares Factor it! = ( x 8 + y 4 )( x 4 + y 2 ) ( x 4 - y 2 ) Factor again! = ( x 8 + y 4 )( x 4 + y 2 )( x 2 +y)( x 2 -y) x 16 - y 8 . Factor this difference of two squares. x 16 - y 8 = ( x 8 + y 4 ) Sum of two squares Does not factor ( x 8 - y 4 )  Difference of two squares Factor it! = ( x 8 + y 4 )( x 4 + y 2 ) ( x 4 - y 2 ) Factor again! = ( x 8 + y 4 )( x 4 + y 2 )( x 2 +y)( x 2 -y)

Finally, the factorization is complete.

These types of products appear from time to time, so be aware that you may have to factor more than once.

Practice Set B

Factor each binomial completely.

Exercise 7

m 4 − n 4 m 4 − n 4

Solution

( m 2 + n 2 )(m−n)(m+n) ( m 2 + n 2 )(m−n)(m+n)

Exercise 8

16 y 8 −1 16 y 8 −1

Solution

(4 y 4 +1)(2 y 2 +1)(2 y 2 −1) (4 y 4 +1)(2 y 2 +1)(2 y 2 −1)

Perfect Square Trinomials

Recall the process of squaring a binomial.

( a b ) 2 = a 2 +2 a b+ b 2 ( a b ) 2 a 2 2 a b b 2 ( a b ) 2 = a 2 −2 a b+ b 2 ( a b ) 2 a 2 2 a b b 2

Table 1
Our Method Is We Notice
Square the first term. The first term of the product should be a perfect square.
Take the product of the two terms and double it. The middle term of the product should be divisible by 2 (since it’s multiplied by 2).
Square the last term. The last term of the product should be a perfect square.

Perfect square trinomials always factor as the square of a binomial.

To recognize a perfect square trinomial, look for the following features:

  1. The first and last terms are perfect squares.
  2. The middle term is divisible by 2, and if we divide the middle term in half (the opposite of doubling it), we will get the product of the terms that when squared produce the first and last terms.

In other words, factoring a perfect square trinomial amounts to finding the terms that, when squared, produce the first and last terms of the trinomial, and substituting into one of the formula

a 2 +2ab+ b 2 = (a+b) 2 a 2 -2ab+ b 2 = (a-b) 2 a 2 +2ab+ b 2 = (a+b) 2 a 2 -2ab+ b 2 = (a-b) 2

Sample Set C

Factor each perfect square trinomial.

Example 6

x 2 +6x+9 x 2 +6x+9 . This expression is a perfect square trinomial. The x 2 x 2 and 9 are perfect squares.

The terms that when squared produce x 2 x 2 and 9 are x x and 3, respectively.

The middle term is divisible by 2, and 6 x 2 =3 x 6 x 2 3 x. The 3 x 3x is the product of x x and 3, which are the terms that produce the perfect squares.

x 2 +6x+9= (x+3) 2 x 2 +6x+9= (x+3) 2

Example 7

x 4 -10 x 2 y 3 +25 y 6 x 4 -10 x 2 y 3 +25 y 6 . This expression is a perfect square trinomial. The x 4 x 4 and 25 y 6 25 y 6 are both perfect squares. The terms that when squared produce x 4 x 4 and 25 y 6 25 y 6 are x 2 x 2 and 5 y 3 5 y 3 , respectively.

The middle term -10 x 2 y 3 -10 x 2 y 3 is divisible by 2 2 . In fact, -10 x 2 y 3 2 =-5 x 2 y 3 -10 x 2 y 3 2 =-5 x 2 y 3 . Thus,

x 4 -10 x 2 y 3 +25 y 6 = ( x 2 -5 y 3 ) 2 x 4 -10 x 2 y 3 +25 y 6 = ( x 2 -5 y 3 ) 2

Example 8

x 2 +10x+16 x 2 +10x+16 . This expression is not a perfect square trinomial. Although the middle term is divisible by 2,? 10x 2 =5x 2,? 10x 2 =5x , the 5 and x x are not the terms that when squared produce the first and last terms. (This expression would be a perfect square trinomial if the middle term were 8x 8x .)

Example 9

4 a 4 +32 a 2 b-64 b 2 4 a 4 +32 a 2 b-64 b 2 . This expression is not a perfect square trinomial since the last term -64 b 2 -64 b 2 is not a perfect square (since any quantity squared is always positive or zero and never negative).

Thus, 4 a 4 +32 a 2 b-64 b 2 4 a 4 +32 a 2 b-64 b 2 cannot be factored using this method.

Practice Set C

Factor, if possible, the following trinomials.

Exercise 9

m 2 -8m+16 m 2 -8m+16

Solution

(m-4) 2 (m-4) 2

Exercise 10

k 2 +10k+25 k 2 +10k+25

Solution

(k+5) 2 (k+5) 2

Exercise 11

4 a 2 +12a+9 4 a 2 +12a+9

Solution

(2a+3) 2 (2a+3) 2

Exercise 12

9 x 2 -24xy+16 y 2 9 x 2 -24xy+16 y 2

Solution

(3x-4y) 2 (3x-4y) 2

Exercise 13

2 w 3 z+16 w 2 z 2 +32w z 3 2 w 3 z+16 w 2 z 2 +32w z 3

Solution

2wz (w+4z) 2 2wz (w+4z) 2

Exercise 14

x 2 +12x+49 x 2 +12x+49

Solution

not possible

Exercises

For the following problems, factor the binomials.

Exercise 15

a2−9 a 2 9

Solution

( a+3 )( a−3 ) ( a+3 )( a−3 )

Exercise 16

a2−25 a 2 25

Exercise 17

x2−16 x 2 16

Solution

( x+4 )( x−4 ) ( x+4 )( x−4 )

Exercise 18

y2−49 y 2 49

Exercise 19

a2−100 a 2 100

Solution

( a+10 )( a−10 ) ( a+10 )( a−10 )

Exercise 20

b2−36 b 2 36

Exercise 21

4 a2−64 4 a 2 64

Solution

4( a+4 )( a−4 ) 4( a+4 )( a−4 )

Exercise 22

2 b2−32 2 b 2 32

Exercise 23

3 x2−27 3 x 2 27

Solution

3( x+3 )( x−3 ) 3( x+3 )( x−3 )

Exercise 24

5 x2−125 5 x 2 125

Exercise 25

4 a2−25 4 a 2 25

Solution

( 2a+5 )( 2a−5 ) ( 2a+5 )( 2a−5 )

Exercise 26

9 x2−100 9 x 2 100

Exercise 27

36 y2−25 36 y 2 25

Solution

( 6y+5 )( 6y−5 ) ( 6y+5 )( 6y−5 )

Exercise 28

121 a2−9 121 a 2 9

Exercise 29

12 a2−75 12 a 2 75

Solution

3( 2a+5 )( 2a−5 ) 3( 2a+5 )( 2a−5 )

Exercise 30

10 y2−320 10 y 2 320

Exercise 31

8 y2−50 8 y 2 50

Solution

2( 2y+5 )( 2y−5 ) 2( 2y+5 )( 2y−5 )

Exercise 32

a2 b2−9 a 2 b 2 9

Exercise 33

x2 y2−25 x 2 y 2 25

Solution

( xy+5 )( xy−5 ) ( xy+5 )( xy−5 )

Exercise 34

x4 y4−36 x 4 y 4 36

Exercise 35

x4 y4−9 a2 x 4 y 4 9 a 2

Solution

( x 2 y 2 +3a )( x 2 y 2 −3a ) ( x 2 y 2 +3a )( x 2 y 2 −3a )

Exercise 36

a2 b4−16 y4 a 2 b 4 16 y 4

Exercise 37

4 a2 b2−9 b2 4 a 2 b 2 9 b 2

Solution

b 2 ( 2a+3 )( 2a−3 ) b 2 ( 2a+3 )( 2a−3 )

Exercise 38

16 x 2 −25 y 2 16 x 2 25 y 2

Exercise 39

a 2 − b 2 a 2 b 2

Solution

( a+b )( a−b ) ( a+b )( a−b )

Exercise 40

a 4 − b 4 a 4 b 4

Exercise 41

x 4 − y 4 x 4 y 4

Solution

( x 2 + y 2 )( x+y )( x−y ) ( x 2 + y 2 )( x+y )( x−y )

Exercise 42

x 8 − y 2 x 8 y 2

Exercise 43

a 8 − y 2 a 8 y 2

Solution

( a 4 +y )( a 4 −y ) ( a 4 +y )( a 4 −y )

Exercise 44

b 6 − y 2 b 6 y 2

Exercise 45

b 6 − x 4 b 6 x 4

Solution

( b 3 + x 2 )( b 3 − x 2 ) ( b 3 + x 2 )( b 3 − x 2 )

Exercise 46

9− x 2 9 x 2

Exercise 47

25− a 2 25 a 2

Solution

( 5+a )( 5−a ) ( 5+a )( 5−a )

Exercise 48

49−16 a 2 49 16 a 2

Exercise 49

100−36 b 4 100 36 b 4

Solution

4( 5+3 b 2 )( 5−3 b 2 ) 4( 5+3 b 2 )( 5−3 b 2 )

Exercise 50

128−32 x 2 128 32 x 2

Exercise 51

x4−16 x 4 16

Solution

( x 2 +4 )( x+2 )( x−2 ) ( x 2 +4 )( x+2 )( x−2 )

Exercise 52

2a b 3 − a 3 b 2a b 3 − a 3 b

Exercise 53

a 4 − b 4 a 4 b 4

Solution

( a 2 + b 2 )( a+b )( a−b ) ( a 2 + b 2 )( a+b )( a−b )

Exercise 54

a 16 − b 4 a 16 b 4

Exercise 55

x 12 − y 12 x 12 y 12

Solution

( x 6 + x 6 )( x 3 + y 3 )( x 3 − y 3 ) ( x 6 + x 6 )( x 3 + y 3 )( x 3 − y 3 )

Exercise 56

a 2 c−9 c a 2 c 9 c

Exercise 57

a 3 c 2 −25a c 2 a 3 c 2 −25a c 2

Solution

a c 2 ( a+5 )( a−5 ) a c 2 ( a+5 )( a−5 )

Exercise 58

a 4 b 4 c 2 d 2 −36 x 2 y 2 a 4 b 4 c 2 d 2 −36 x 2 y 2

Exercise 59

49 x 2 y 4 z 6 −64 a 4 b 2 c 8 d 10 49 x 2 y 4 z 6 −64 a 4 b 2 c 8 d 10

Solution

( 7x y 2 z 3 +8 a 2 b c 4 d 5 )( 7x y 2 z 3 −8 a 2 b c 4 d 5 ) ( 7x y 2 z 3 +8 a 2 b c 4 d 5 )( 7x y 2 z 3 −8 a 2 b c 4 d 5 )

For the following problems, factor, if possible, the trinomials.

Exercise 60

x 2 +8 x+16 x 2 8 x 16

Exercise 61

x 2 +10 x+25 x 2 10 x 25

Solution

( x+5 ) 2 ( x+5 ) 2

Exercise 62

a 2 +4 a+4 a 2 4 a 4

Exercise 63

a 2 +12 a+36 a 2 12 a 36

Solution

( a+6 ) 2 ( a+6 ) 2

Exercise 64

b 2 +18 b+81 b 2 18 b 81

Exercise 65

y 2 +20 y+100 y 2 20 y 100

Solution

( y+10 ) 2 ( y+10 ) 2

Exercise 66

c 2 +6 c+9 c 2 6 c 9

Exercise 67

a 2 −4 a+4 a 2 4 a 4

Solution

( a−2 ) 2 ( a−2 ) 2

Exercise 68

b 2 −6 b+9 b 2 6 b 9

Exercise 69

x 2 −10 x+25 x 2 10 x 25

Solution

( x−5 ) 2 ( x−5 ) 2

Exercise 70

b 2 −22 b+121 b 2 22 b 121

Exercise 71

a 2 −24 a+144 a 2 24 a 144

Solution

( a−12 ) 2 ( a−12 ) 2

Exercise 72

a 2 +2 a+1 a 2 2 a 1

Exercise 73

x 2 +2 x+1 x 2 2 x 1

Solution

( x+1 ) 2 ( x+1 ) 2

Exercise 74

x 2 −2 x+1 x 2 2 x 1

Exercise 75

b 2 −2 b+1 b 2 2 b 1

Solution

( b−1 ) 2 ( b−1 ) 2

Exercise 76

4 a 2 +12 a+9 4 a 2 12 a 9

Exercise 77

9 x 2 +6 x+1 9 x 2 6 x 1

Solution

( 3x+1 ) 2 ( 3x+1 ) 2

Exercise 78

4 x 2 +28 x+49 4 x 2 28 x 49

Exercise 79

16 a 2 −24 a+9 16 a 2 24 a 9

Solution

( 4a−3 ) 2 ( 4a−3 ) 2

Exercise 80

25 a 2 −20 a+4 25 a 2 20 a 4

Exercise 81

9 x 2 +6 x y+ y 2 9 x 2 6 x y y 2

Solution

( 3x+y ) 2 ( 3x+y ) 2

Exercise 82

16 x 2 +24 x y+9 y 2 16 x 2 24 x y 9 y 2

Exercise 83

36 a 2 +60 a b+25 b 2 36 a 2 60 a b 25 b 2

Solution

( 6a+5b ) 2 ( 6a+5b ) 2

Exercise 84

4 x 2 −12 x y+9 y 2 4 x 2 12 x y 9 y 2

Exercise 85

12 a 2 −60 a+75 12 a 2 60 a 75

Solution

3 ( 2a−5 ) 2 3 ( 2a−5 ) 2

Exercise 86

16 x 2 +8 x+1 16 x 2 8 x 1

Exercise 87

32 x 2 +16 x+2 32 x 2 16 x 2

Solution

2 ( 4x+1 ) 2 2 ( 4x+1 ) 2

Exercise 88

x 2 +x+1 x 2 x 1

Exercise 89

4 a 2 +a+9 4 a 2 a 9

Solution

not factorable

Exercise 90

9 a 2 −21 a+49 9 a 2 21 a 49

Exercise 91

x 5 +8 x 4 +16 x 3 x 5 8 x 4 16 x 3

Solution

x 3 ( x+4 ) 2 x 3 ( x+4 ) 2

Exercise 92

12 a 3 b-48 a 2 b 2 +48a b 3 12 a 3 b-48 a 2 b 2 +48a b 3

EXERCISES FOR REVIEW

Exercise 93

((Reference)) Factor ( m−3 ) x−( m−3 ) y ( m 3 ) x ( m 3 ) y .

Solution

( m−3 )( x−y ) ( m−3 )( x−y )

Exercise 94

((Reference)) Factor 8 x m+16 x n+3 y m+6 y n 8x m 16 x n 3 y m 6 yn by grouping.

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