# Factoring Polynomials

## III. Factor by Grouping(4 Terms)

**Strategy:** Group the four terms in pairs and find
the GCF of each pair. Both pairs should then have

a common __binomial__ factor.

**Example 1:**Factor 3ax-3ay-2bx+2by

__3ax-3ay__-__2bx+2by __

GCF of 1^{st} group is 3a

GCF of 2^{nd} group is -2b

(Note: If the two groups are separated by a subtraction always factor out a negative GCF from the

second group.)

Factoring each group yields __3a(x-y)__-__2b(x-y)__

Both have a common binomial factor (x-y) which can be factored out leaving the other factor(3a-2b).

**Solution: ** (x-y)(3a-2b)

**Example 2: ** Factor 4X^{2}-25X-21

**Strategy:** Notice!...this polynomial is the same as
in section II and it doesn't have 4 terms. But,

if we replace the middle term-25x with the factor pair of
**ac** from the list that adds to -25 we can

write this trinomial with 4 terms and solve by grouping .This is an alternative way of factoring a

__trinomial__ where** a≠1.**

__4x ^{2}-28x__+

__3x-21__

GCF of 1

^{st}group is 4x

GCF of 2

^{nd}group if 3

Factoring each group yields
__4x(x-7)__+__3(x-7)__

Both have a common binomial factou (x-7) which can be factored out leaving the other factor(4x-3)

**Solution:** (x-7)(4x-3)

## Ⅳ. Special Cases (2 Terms)

** i) Difference of Squares**

**Strategy: **Check the binomial to see if it is a
difference of squares.

A Difference of Squares** A ^{2}-B^{2}**
factors into

**(A+B)(A-B)**

TEST:

i) **1 ^{st} and last terms **must be perfect squares.

ii) Must be a

__subtraction.__

**NOTE: **Besides a possible common factor, the Sum of
Squares **A ^{2} + B^{2} **is PRIME

**(not factorable)**.

**Example:** Factor 9x^{2}-16

Test:

i) **1 ^{st} term** is 9x

^{2}

**→**(3x)

^{2}and

**last term,**is 16

**→**(4)

^{2},both perfect squares with roots 3x and 4 as shown.

ii) are we subtracting? YES!

The binomial passes both test. It is a Difference of
Squares of the form

**A ^{2}-B^{2} = (A+B)(A-B)** where 3x

**→**

**A**and 4

**→**

**B**

**Solution:** (3x+4)(3x-4)

**ii) Difference OR Sum of Cubes**

**Strategy:** Check the binomial to see if it is a
difference of cubes or a sum of cubes.

A Difference of Cubes **A3-B3** factors into **
(A-B)(A ^{2}+AB+B^{2})**

A Sum of Cubes

**A3+B3**factor into

**(A+B)(A**

^{2}-AB+B^{2})Note: The trinomial factor is often PRLME and cannot be factored further.

TEST:

i) **1 ^{st} and last terms** must be perfect cubes.

ii) Can be erther a

__subtraction__

**OR**an

__addition.__

**Example 1:** Factor 27x^{3}+8

Test:

i) **1 ^{st} term** is 27x

^{3}

**→**(3x)

^{3}and

**last term**is 8

**→**(2)

^{3},both perfect cubes with roots 3x and 2 as shown.

ii) it is the

**sum**of cubes.

The binomial is a __Sum__** **of Cubes of the form
**A ^{3}+B^{3} = (A+B)(A^{2}-AB+B^{2})**

where 3x

**→**

**A**and 2

**→**

**B**

**Solution: **(3x+2)(9z^{2}-6x+4)

**Example 2:** Factor 1-64x^{3}

Test:

i) **1 ^{st} term** is 1

**→**(1)

^{3}and

**last term**is 64x

^{3}

**→**(4x)

^{3}, both perfect cubes with roots 1 and 4x as shown.

ii) it is the

**difference**of cubes.

The binomial is a__ Difference __of Cubes of the form
**A ^{3}-B^{3} = (A-B)(A^{2}+AB+B^{2})**

where 1

**→**

**A**and 4x

**→**

**B**

**Solution:** (1-4x)(1+4x+16x^{2})

## WORKED EXAMPLES

1. Factor 16x^{4}-1

(4x^{2})^{2}-(1)^{2} Difference of
Squares

(4x^{2}+1)(4x^{2}-1) Another Difference of Squares!

(4x^{2}+1)((2x)^{2}-(1)^{2})

(4x^{2}+1)(2x+1)(2x-1)

2. Factor 3x^{2}+27

3(x^{2}+9)

GCF

(X^{2}+9) can NOT be factored further it is the Sum of __Squares__
which is prime.

3. Factor 8a^{3}+27b^{3}

(2a)^{3}+(3b)^{3} Sum of __Cubes__

(2a+3b)(4a^{2}-6ab+9b^{2})

4. Factor 2x^{2}-xy-6y^{2}

(_x_y)(_x_y) Factor pairs of ac = -12 that add to the
middle are 3 and -4

**Practice Exercises**

**Answers to Practice Ecercises**