37

CHAPTER 2 POLYNOMIAL FACTORING

2.1 Factoring out the GCF and Factoring by Grouping To factor out the greatest common factor (GCF), simply apply the distributive property of multiplication over addition (or subtraction) in reverse as follows:

( )ab ac a b c+ = + .

Following are some examples to review the concept. Examples

1. Factor: 5y yb yz− +

Solution: The factor y is common to the three terms so factoring it out gives

( )5 5y yb yz y b z− + = − + .

2. Factor: 2 3 4 3 5 324 12 20 4x y xy x y xy− + +

Solution: To get the GCF we can do something like the following:

2 3

4

3 5

3

3

24 2 2 2 3

12 2 2 3

20 5 2 2

4 2 2

GCF 2 2 4

x y x x y y y

xy x y y y y

x y x x x y y y y y

xy x y y y

x y y y xy

= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ ⋅ =

Thus, ( )2 3 4 3 5 3 3 2 224 12 20 4 4 6 3 5 1x y xy x y xy xy x y x y− + + = − + + .

3. ( ) ( )3 8a x y b x y+ − +

Solution: The binomial x y+ is common to the two “terms” and factoring it out

gives ( ) ( ) ( )( )3 8 3 8a x y b x y x y a b+ − + = + − .

4. ( ) ( )6 11n a b c y a b c+ + − + +

Solution: The trinomial a b c+ + is common to the two “terms” and factoring it

out gives ( ) ( ) ( )( )6 11 6 11n a b c y a b c a b c n y+ + − + + = + + − .

5. ( ) ( ) ( )( ) ( )( )2 3 2 3 5 2 4 2x w p y w p z w p+ − + − + − + + − − +

Solution: The trinomial 2w p− + is common to the three “terms”, thus

( )( ) ( )( ) ( )( )

( ) ( ) ( ) ( )

( )( )

( )( )

2 3 2 3 5 2 4 2

2 2 3 3 5 4

2 2 3 3 5 4

2 2 3 6

x w p y w p z w p

w p x y z

w p x y z

w p x y z

+ − + − + − + + − − +

= − + + − + + −

= − + + − − + −

= − + − + −

38

6. ( ) ( )2 5 5t m t− + −

Solution: Note that 5 and 5t t− − are opposites of each other so that one can

be written as the negative of the other, i.e., for example we can have

( )5 5t t− = − −

so that

( ) ( )

( ) ( )

( )( )

2 5 5

2 5 5

5 2

t m t

t m t

t m

− + −

= − − −

= − −

7. ( ) ( )2 24 2 3 8 3 2m n y mn y− − −

Solution: We can again write one as the opposite of the other to get

( ) ( )

( ) ( )

( )( )

( )( )

2 2

2 2

2 2

4 2 3 8 3 2

4 2 3 8 2 3

4 8 2 3

4 2 2 3

m n y mn y

m n y mn y

m n mn y

mn m n y

− − −

= − + −

= + −

= + −

Sometimes if we consider all the terms we will not see any common factor but if we group terms we are able to factor out some common factors. This method is called factoring by grouping.

8. 5 5x y mx my+ + +

Solution: Note that the first two terms as well as the last two terms have a common factor so we can group them to get

( ) ( )

( ) ( )

( ) ( )

5 5

5 5

5

5

x y mx my

x y mx my

x y m x y

x y m

+ + +

= + + +

= + + +

= + +

9. 3 22 8 4 16a a a− − +

Solution: We can group the first two and the last two:

( ) ( ) ( )

( ) ( )

( )( )

3 2

3 2

2

2

Remember to change the signs inside the for the second group!

Factor out the GCF from each group.

Factor out the binomial GCF.

2 8 4 16

2 8 4 16

2 4 4 4

2 4 4

2

a a a

a a a

a a a

a a

a

− − +

= − − −

= − − −

= − −

= ( )( )2Factor out the GCF 2 from the first factor.2 4 a− −

39

2.2 Factoring Quadratic Trinomials

In this section we shall review how to factor trinomials of the form 2ax bx c+ +

or what I call trinomial squares. We first consider the case where the leading

coefficient a is 1, i.e. 2x bx c+ + . Recall that this factors into two

binomials ( )( )x m x n+ + , where m and n are factors of c that add up to b.

Examples

1. Factor: 2 14 40x x+ +

Solution: The factors of 40 that add up to 14 are 10 and 4, thus

( )( )2 14 40 10 4x x x x+ + = + + .

2. Factor: 2 11 28y y− +

Solution: The factors of 28 that add up to 11− are 4 and 7− − , thus

( )( )2 11 28 4 7y y y y− + = − − .

Recall that a second variable may be present but the technique stays the same.

3. Factor: 2 22 35a ab b+ −

Solution: The factors of 35− that add up to 2 are 7 and 5− , thus

( )( )2 22 35 7 5a ab b a b a b+ − = + − .

4. Factor: 2 26 27m mn n− −

Solution: The factors of 27− that add up to 6− are 9 and 3− , thus

( )( )2 26 27 9 3m mn n m n m n− − = − + .

You may want to use the following diagram to find the factors of the constant term c that add up to the middle coefficient b: Next, let us consider the case when the leading coefficient is not equal to 1,

i.e., trinomials of the form 2ax bx c+ + . Recall that this can be factored by

doing trial-and-error methods, by grouping, or by the bottoms-up method. Let us review the last two methods. Both grouping and bottoms-up methods start out the same way as the case where the leading coefficient is 1, i.e. we also want to find two factors of a number that add up to another number. The only difference is that we consider

+ b

i c

m n

40

the factors of the product ac rather than of c alone. Following are the steps involved in the grouping method:

a. Find the factors of ac that add up to b. b. Rewrite the middle term using the factors found in step a. c. Factor the resulting polynomial with four terms by grouping.

Following are the steps involved in the bottoms-up method:

a. Find two factors of ac that add up to b. b. Divide these two numbers by a. c. Reduce the resulting fractions to lowest terms. d. Do bottoms-up!

Examples

5. Factor: 210 11 3x x+ +

Grouping Method:

a. Find the factors of 10 3⋅ or 30 that add up to 11. These are 5 and 6.

b. Write 11x as 5 6x x+ .

c. Factor 210 5 6 3x x x+ + + by grouping:

( ) ( ) ( ) ( ) ( )( )210 5 6 3 5 2 1 3 2 1 2 1 5 3x x x x x x x x+ + + = + + + = + +

Bottoms-up Method:

a. Find the factors of 10 3⋅ or 30 that add up to 11. These are 5 and 6.

b. Divide 5 and 6 by 10 to get the fractions 5 6

and 10 10

.

c. Reduce each fraction to get 1 3

and 2 5

.

d. Bottoms-up produces the factors ( )( )2 1 5 3x x+ + . Check by multiplying!

6. Factor: 221 2 8y y+ −

Grouping Method:

a. Find factors of ( )21 8⋅ − that add up to 2. Now this product is a pretty big

number. We can write this product as 3 7 2 2 2− ⋅ ⋅ ⋅ ⋅ and let’s “play” with

these factors to find the combination. With some patience one gets

numbers ( ) ( )14 from 7 2 and 12 from 3 2 2⋅ − − ⋅ ⋅ to be the

that will give 2!

b. Write 2y as 14 12y y− .

c. Factor 221 14 12 8y y y+ − − by grouping:

( ) ( ) ( ) ( ) ( )( )221 14 12 8 7 3 2 4 3 2 3 2 7 4y y y y y y y y+ − + = + − + = + −

Bottoms-up Method: a. Same as in the grouping method.

b. Divide 14 and 12− by 21 to get the fractions 14 12

and 21 21

− .

c. Reduce each fraction to get 2 4

and 3 7

− .

d. Bottoms-up produces the factors ( ) ( )3 2 7 4y y+ − .

41

There could be a second variable.

7. Factor: 2 248 26 35n np p− −

Grouping Method:

a. Find factors of ( )48 35⋅ − that add up to 26− . Play with

2 2 2 2 3 5 7− ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ to get ( )56 from 2 2 2 7− ⋅ ⋅ ⋅ and 30 ( )from 2 3 5⋅ ⋅ .

b. Write 26np− as 56 30np np− + .

c. Factor 2 248 56 30 35n np np p− + − by grouping:

( ) ( ) ( ) ( ) ( )( )2 248 56 30 35 8 6 7 5 6 7 6 7 8 5n np np p n n p p n p n p n p− + − = − + − = − +

Bottoms-up Method: a. Same as above.

b. Divide 56 and 30− by 48 to get 56 30

and 48 48

− .

c. Reduce each to get 7 5

and 6 8

− .

d. Bottoms-up produces the factors ( )( )6 7 8 5n p n p− + .

Again, you may want to use the following diagram: Let us redo Examples 5 – 7:

5. Factor: 210 11 3x x+ +

Solution:

Thus, the factored form is ( )( )2 1 5 3x x+ + .

+ b

i ac

m n

+ 11

i 3⋅10

= �6

5 2 3⋅ ⋅

5 1

2=

10

6 3

5=

10

42

6. Factor: 221 2 8y y+ −

Solution:

Thus, the factored form is ( ) ( )3 2 7 4y y+ − .

7. Factor: 2 248 26 35n np p− −

Solution:

Thus, the factored form is ( ) ( )6 7 8 5n n− + .

Warning! It is important to watch out for GCFs before applying bottoms-up!

8. 22 18 40x x− +

Solution: If we did not notice that 2 is the GCF and applied bottoms-up we would do the following:

a. Find factors of 2 40⋅ that add up to 18− : these are 8 and 10− − .

b. Divide 8 and 10− − by 2 to get 4 and 5− − .

Then we get the factored form ( )( )4 5x x− − which of course is WRONG!

We can fix the problem by putting the GCF 2 in the factored form to get

( )( )2 4 5x x− − which is the correct factored form.

Thus, it is a good idea to watch out for any GCF first or else, make sure to check the factored form and put the GCF if necessary.

+ 2

i ( )8⋅ −21

= 3 7 4 2− ⋅ ⋅ ⋅

14 2

3=

21

12 4

7

−= −

21

+ 26−

i ( )35⋅ −48

= 6 8 7 5− ⋅ ⋅ ⋅

56 7

6

−= −

48

30 5

8=

48

43

2.3 Factoring Special Binomials

To factor the difference of squares 2 2a b− we use the formula

( )( )2 2a b a b a b− = + − .

Examples

1. Factor: 2 81x −

Solution: ( )( )2 2 281 9 9 9x x x x− = − = + −

2. Factor: 2 216 49m n−

Solution: ( ) ( ) ( )( )2 22 216 49 4 7 4 7 4 7m n m n m n m n− = − = + −

3. Factor: 2 29 16

4 81a b−

Solution:

2 2

2 29 16 3 4 3 4 3 4

4 81 2 9 2 9 2 9a b a b a b a b

− = − = + −

In general we cannot factor the sum of squares such as 2 21, 4, etc.a x+ + .

We will see later that there are some forms of the sum of squares that can actually be factored. To factor the sum or difference of cubes we use the following formulas:

( )( )

( )( )

3 3 2 2

3 3 2 2

a b a b a ab b

a b a b a ab b

+ = + − +

− = − + +

4. Factor: 3 64x +

Solution: ( )( )3 3 3 264 4 4 4 16x x x x x+ = + = + − +

5. Factor: 3 327 125x y+

Solution: ( ) ( ) ( )( )3 33 3 2 227 125 3 5 3 5 9 15 25x y x y x y x xy y+ = + = + − +

9. Factor: 3 3

27 1000

b c−

Solution:

3 33 3 2 2

27 1000 3 10 3 10 9 30 100

b c b c b c b bc c − = − = − + +

There are occasions when we can apply these factoring techniques more than once within a problem.

10. Factor: 4 16x −

Solution: ( ) ( )( ) ( )( )( )2

4 2 2 2 2 216 4 4 4 2 2 4x x x x x x x− = − = − + = + − +

44

2.4 Factoring by Substitution The method of substitution can be used to reduce a polynomial to a form that can be easily factored. Examples

1. Factor: 4 23 2x x− +

Solution: Let 2u x= . Then

( )

( )( )

( )( ) ( )( )( )

24 2 2 2

2

2 2 2

3 2 3 2

3 2 2 1

2 1 2 1 1

x x x x

u u u u

x x x x x

− + = − +

= − + = − −

= − − = − − +

2. Factor: 4 26 2y y+ −

Solution: Let 2x y= . Then

( )

( )( )

( )( )

24 2 2 2

2

2 2

6 2 6 2

6 2 3 2 2 1

3 2 2 1

y y y y

x x x x

y y

+ − = + −

= + − = + −

= + −

3. Factor: ( ) ( )2

1 4 1 60x x− − − −

Solution: Let 1a x= − . Then

( ) ( )

( )( )

( )( ) ( )( )

2

2

1 4 1 60

4 60 10 6

1 10 1 6 11 5

x x

a a a a

x x x x

− − − −

= − − = − +

= − − − + = − +

4. Factor: 4 416 81x y−

Solution: Let 2 24 and 9u x v y= = . Then

( ) ( )

( )( )

( )( ) ( )( )( )

2 24 4 2 2

2 2

2 2 2 2 2 2

16 81 4 9

4 9 4 9 2 3 2 3 4 9

x y x y

u v u v u v

x y x y x y x y x y

− = −

= − = − +

= − + = − + +

5. Factor: ( ) ( )2 2

1 2a b− − −

Solution: Let 1 and 2x a y b= − = − . Then

( ) ( )

( )( )

( ) ( ) ( ) ( ) [ ][ ]

( )( )

2 2

2 2

1 2

1 2 1 2 1 2 1 2

1 3

a b

x y x y x y

a b a b a b a b

a b a b

− − −

= − = − +

= − − − − + − = − − + − + −

= − + + −

45

2.5 Miscellaneous Factoring Techniques

Factorable Sums of Squares Although we tend to think that sums of squares cannot be factored over the rational numbers, there are actually certain sums of squares that can be

factored! Consider 4 44x y+ . If we add and subtract the expression 2 24x y ,

we will get 4 2 2 4 2 24 4 4x x y y x y+ + − . We can then factor this polynomial by

grouping the first three terms and using the substitution technique (or not ☺):

( ) ( ) ( )

( )( )

2 24 2 2 4 2 2 2 2

2 2 2 2

4 4 4 2 2

2 2 2 2 factoring as the difference of squares

x x y y x y x y xy

x y xy x y xy

+ + − = + −

= + − + +

Special Trinomial Factoring

Consider the trinomial 4 2 2 425x x y y+ + . We may conclude at first that this

trinomial which is in quadratic form cannot be factored since there are no factors of 25 that will give 1 when added. We have, however, the following

situation. Let us add and subtract 2 29x y to get 4 2 2 4 2 210 25 9x x y y x y+ + − .

The resulting polynomial with four terms can then be factored by grouping as shown below and thus is factorable!

( )

( ) ( )

( ) ( )

4 2 2 4 2 2

2 22 2

2 2 2 2

10 25 9

5 3

5 3 5 3

x x y y x y

x y xy

x y xy x y xy

+ + −

= + −

= + − + +

The polynomial 6 6x y− is an interesting factoring problem. One can think of

the polynomial as a difference of squares ( ) ( )2 2

3 3x y− or as a difference of

cubes ( ) ( )3 3

2 2x y− . If we factor it as a difference of squares, we will get:

( ) ( ) ( )( )

( )( )( )( )

2 23 3 3 3 3 3

2 2 2 2 .

x y x y x y

x y x xy y x y x xy y

− = − +

= − + + + − +

If we factor it as a difference of cubes, we will get:

( ) ( ) ( )( ) ( )( )( )3 3

2 2 2 2 4 2 2 4 4 2 2 4x y x y x x y y x y x y x x y y− = − + + = − + + + .

Which is the complete factored form? It seems that factoring 6 6x y− as a

difference of squares is the better option. Let us reconcile the answers. If we

consider the trinomial factor 4 2 2 4x x y y+ + , it looks like it might be treated

similarly as the example we explained above. Note that if we add and subtract 2 2x y to the trinomial, then we will get 4 2 2 4 2 22x x y y x y+ + − which can be

factored by grouping as follows:

( ) ( ) ( ) ( )( )2 24 2 2 4 2 2 2 2 2 2 2 22x x y y x y x y xy x y xy x y xy+ + − = + − = + − + + .

Note that these trinomials are the same trinomials we obtained in the first method of factoring that we used.

46

2.6 Factoring Using Combined Techniques Following is a general strategy for factoring polynomials: 1. Factor out any GCFs. 2. Count the terms. a. If there are 2 terms, check if the polynomial looks like

i. The difference of squares 2 2a b− ⇒ factor as ( )( )a b a b− + .

ii. The sum of cubes 3 3a b+ ⇒ factor as ( )( )2 2a b a ab b+ − + .

iii. The difference of cubes 3 3a b− ⇒ factor as ( )( )2 2a b a ab b− + + .

b. If there are 3 terms, check if the polynomial looks like 2ax bx c+ + , then

i. If 1a = , find the factors of c that add up to b, say m and n, then the

factors are ( )( )x m x n+ + .

ii. If 1a ≠ , you can do

(1) trial-and-error method (2) grouping method (3) bottoms-up method where you find the factors of ac that add up

to b, divide these by a, reduce to lowest terms, say and m p n q

and you get the factored form ( )( )px m qx n+ + .

c. If there are 4 or more terms, try factoring by grouping. It is better to do grouping first before factoring out any GCFs when there are 4 or more terms. Note: Watch out for the special cases explained in the previous section. Examples

1. Factor: 3 212 27a ab−

Solution:

( )

( ) ( )

3 2 2 212 27 3 4 9 Factor out the GCF.

3 2 3 2 3 Factor the difference of squares.

a ab a a b

a a b a b

− = −

= + −

3. Factor: 440 5x x−

Solution:

( )

( )( )

4 3

2

40 5 5 8 1 Factor out the GCF.

5 2 1 4 2 1 Factor the difference of cubes.

x x x x

x x x x

− = −

= − + +

4. Factor: 3 22 4 18 36y y y+ − −

Solution: 3 22 4 18 36y y y+ − −

( ) ( )

( ) ( )

( )( )

( )( )

3 2

2

2

2

2 4 18 36 Group the 1st two and the last two terms.

2 2 18 2 Factor out the GCF from each group.

2 2 18 Factor out the binomial GCF.

2 2 9

y y y

y y y

y y

y y

= + − +

= + − +

= + −

= + −

( )( )( )

Factor out the GCF 2 from the 2nd factor.

2 2 3 3 Factor the difference of squares.y y y= + + −

47

5. Factor: 2 22 1m m n− + − Solution:

( )

( )

( )( )

2 2

2 2

2 2

2 1

2 1 Group the first three terms.

1 Factor the perfect square trinomial.

1 1 Factor the difference of squares.

m m n

m m n

m n

m n m n

− + −

= − + −

= − −

= − + − −

6. Factor: 2 2 26 9 2a a b bc c− + − − −

Solution:

( ) ( )

( ) ( )

( )( )

2 2 2

2 2 2

2 2

6 9 2

6 9 2 Group the 1st three and last three terms.

3 Factor the perfect square trinomials.

3 3 Factor the difference of squ

a a b bc c

a a b bc c

a b c

a b c a b c

− + − − −

= − + − + +

= − − +

= − + + − − − ares.*

*You may want to use a substitution to convince yourself that the second to the last expression is a difference of squares.

7. Factor: 3 2 3 25 40 30 240pm p p m p− − +

Solution:

( ) ( )

( ) ( )

( )( )

3 2 3 2

3 2 3 2

3 2 3

2 3

5 40 30 240

5 40 30 240 Group the 1st two and last two terms.

5 8 30 8 Factor out the GCFs from each group.

5 30 8 Factor out the

pm p p m p

pm p p m p

p m p m

p p m

− − +

= − − −

= − − −

= − −

( )( )( )2

binomial GCF.

5 1 6 2 2 4 Factor out the GCF from the 1st factor;

factor the difference of cubes.

p p m m m= − − + +

8. Factor: ( ) ( )2

2 22 14 2 15x x x x+ − + −

Solution: We can first let 2 2x x y+ = and factor the polynomial

2 14 15y y− − into ( ) ( )15 1y y− + . Then replacing y by the original expression,

we get ( )( )2 22 15 2 1x x x x+ − + + and we can factor each quadratic trinomial

factor to get the complete factored form ( )( )( )2

5 3 1x x x+ − + .

48

2.7 Chapter Review

GCF Factoring

( )3 2 4 2 320 15 5 4 3x y x y x y x y− = −

( ) ( ) ( )( )3 4 5 4 4 3 5a z b z z a b− − − = − −

Factoring the Difference of Squares

( ) ( ) ( )( )222 4 2 2 216 49 4 7 4 7 4 7x y x y x y x y− = − = − +

( ) ( ) ( ) ( )( )2 2 2225 4 5 4 5 4 5 4a b c a b c a b c a b c− + = − + = − − + +

Factoring the Sum or Difference of Cubes

( ) ( ) ( )( )333 6 2 2 2 2 48 125 2 5 2 5 4 10 25n p n p n p n np p+ = + = + − +

3 39 12 3 4 3 4 6 3 4 88 2 2 2 4

27 343 3 7 3 7 9 21 49

a b a b a b a a b b − = − = − + +

Factoring by Grouping

( ) ( ) ( ) ( )

( )( ) ( )( )( )( )

5 2 3 5 2 3 2 3 3

3 2 2

9 9 9 9 1 9 1

1 1 1 1 1 1

a a a a a a a a a

a a a a a a a

− − + = − − − = − − −

= − − = − + + − +

( ) ( ) ( )

( )( )

2 22 2 2 24 6 9 4 6 9 2 3

2 3 2 3

x y y x y y x y

x y x y

− − − = − + + = − +

= − − + +

Factoring Quadratic Trinomials

( )( )

( )( )

( ) ( )( ) ( )( )( )

( ) ( ) ( )( ) ( ) ( )

2

2 2

24 2 2 2 2 2 2

2

6 40 10 4

6 40 10 4

6 40 6 40 10 4 10 2 2

1 6 1 40 1 10 1 4 9 5

a a a a

a ab b a b a b

y y y y y y y y y

x x x x x x

+ − = + −

+ − = + −

+ − = + − = + − = + − +

− + − − = − + − − = + −

( )( )

( ) ( )

( ) ( )( )

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

( )( ) ( )( )

2

2 2

26 3 3 3 3 3

2 2

8 14 15 4 3 2 5

8 14 15 4 3 2 5

8 14 15 8 14 15 4 3 2 5

8 2 14 2 1 15 1 4 2 3 1 2 2 5 1

4 8 3 3 2 4 5 5 4 3 11 2 5 1

a a a a

a ab b a b a b

x x x x x x

a a b b a b a b

a b a b a b a b

+ − = − +

+ − = − +

+ − = + − − +

− + − + − + = − − + − + +

= − − − − + + = − − + +

Miscellaneous Factoring

( ) ( ) ( ) ( )( )24 4 2 2 2 2 264 16 64 16 8 4 4 8 4 8a a a a a a a a a a+ = + + − = + − = − + + +

( ) ( ) ( )

( )( )

2 24 2 2 4 4 2 2 4 2 2 2 2

2 2 2 2

9 25 10 25 5

5 5

x x y y x x y y x y x y xy

x xy y x xy y

+ + = + + − = + −

= + + − +