When factoring is completed:

1) Check to see if one of the polynomial factors can be factored again.

2) Double check for other GCFs (greatest common factors)

> A factor that __every__ term in the expression has in
common

> Reverse distribution

**Examplea:** To find the GCF for 12x^{4} - 6x^{3}+ 3x^{2}

12x^{4}-6x^{3}+3x^{2} means (2·2·3·x·x·x·x)-(2·3·x·x·x)+(3·x·x)

all three terms have 3 and x·x as common factors

So the GCG is** 3x ^{2}**

Note:

1) 3 is the greatest common numerical factor because it is the largest number
that divides into every term evenly.

2) x^{2} is the greatest common variable factor because it is the variable to the
largest power that divides evenly into all three terms.

To factor 12x^{4}-6x^{3}+3x^{2} thi9ne reverse distribution **
3x ^{2}(4x^{2}-2x+1)**

**i) Special Case (Perfect Square Trinomial)**

**Strategy:** Test the trinomial to see if it satisfies the conditions of a
Perfect Square Trinomial.

A Perfect Square Trinomial:

**A ^{2}+2AB+B^{2} **factors into

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

ii)

**Example:** Factor 9x^{2}-24x+6

Test:

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

ii) does the

The trinomial passes both test. It is a Perfect Square
Triniomial of the type

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

**Solution:**(3x-4)^{2}

**ii) For a = 1 (lead coefficient is 1)**

**Strategy:** Find the product ac then determine which
factor pairs of ac add up to b.

**Ecample: **Factor x^{2}-x-12 Note: a=1,b=-1,and c=-12

ac==>(1)(-12)=-12=ac

Note:

i) You don’t have to list all
factor pairs, just find the two that add to b.

ii) If there are no factor pairs that add to b then the polynomial is "prime".

Solution must be of the form (x _)(x _) where the factor pair and corresponding positive or negative sign supply the missing terms.

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

**iii) For a≠1**

**Strategy**: Find the product ac then determine the
factor pairs of ac that add up to b.

**Examole:** Factor 4x^{2}-25x-21 Note: a=4.b=-25,c=-21

Note: i) You don't have to list all factor pairs, just find the two shat add to b.

ii) If there are no factor pairs that add to b then the polynomial is " prime".

Solution must be of the form (_x _)(_x _)

Clues to help you find the missing coefficients and terms:

i) List the factor pairs of **a** to find possible
coefficients of x.

ii) List the factor pairs of **c** to possible missing final terms.

iii)Product of "inners"
and "outers" must be from the list of factor pairs of ac above that
to **b**.

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

**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__+

GCF of 1

GCF of 2

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)

** 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

TEST:

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

ii) Must be a

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

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

Test:

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

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

**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

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

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

Test:

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

ii) it is the

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

where 3x

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

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

Test:

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

ii) it is the

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

where 1

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

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**