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# Factoring Help

based on 1 rating
By McGraw-Hill Professional
Updated on Sep 26, 2011

## Factoring Using the Distributive Property

The distributive property, a(b + c) = ab + ac , can be used to factor a quantity from two or more terms. In the formula ab + ac = a(b + c), a is factored from (or divided into) ab and ac . The first step in factoring is to decide what quantity you want to factor from each term. Second write each term as a product of the factor and something else (this step will become unnecessary once you are experienced). Third apply the distribution property in reverse.

#### Examples

4 + 6x

Each term is divisible by 2, so factor 2 from 4 and 6 x : 4 + 6 x = 2 · 2 + 2 · 3 x = 2(2 + 3 x ).

2x + 5x2 = x · 2 + x · 5x = x (2 + 5x )

3x2 + 6x = 3x · x + 3x · 2 = 3x (x + 2)

8x + 8 = 8 · x + 8 · 1 = 8(x + 1)

4xy + 6x2 + 2xy2 = 2x · 2y + 2x · 3x + 2x · y2 = 2x (2y + 3x + y2)

### Factoring Complicated Expressions

Complicated expressions can be factored in several steps. Take for example 48x5y3z6 + 60x4yz3 + 36x6y2z , each term is divisible by 12xyz . Start with this.

48x5y3z6 + 60x4yz3 + 36x6y2z

= 12xyz · 4x4y2z5 + 12xyz · 5x3z2 + 12xyz · 3x5y

= 12xyz (4x4y2z5 + 5x3z2 + 3x5y )

Each term in the parentheses is divisible by x2 :

4x4y2z5 + 5x3z2 + 3x5y

= x2 · 4x2y2z5 + x2 · 5xz2 + x2 · 3x3y

= x2(4x2y2z5 + 5xz2 + 3x3y)

48x5y3z6 + 60x4yz3 + 36x6y2z

= 12xyz · x2 (4x2y2z5 + 5xz2 + 3x3y)

= 12x3yz (4x2y2z5 + 5xz2 + 3x3y)

Find practice problems and solutions at Factoring Practice Problems - Set 1.

## Factoring Negative Quantities

Factoring a negative quantity has the same effect on signs within parentheses as distributing a negative quantity does—every sign changes. Negative quantities are factored in the next examples and practice problems.

#### Examples

 x + y = –(– x – y ) –4 + x = –(4 – x) –2 – 3x = –(2 + 3x) 2x2 + 4x = –2 x (– x – 2) –14xy + 21x 2y = –7xy (2 – 3x) 12xy – 25x = – x (–12y + 25) 16y2 – 1 = –(–16y2 + 1) = –(1 – 16y2) –4x + 3y = –(4x – 3y) x – y – z + 5 = –(– x + y + z – 5)

Find practice problems and solutions at Factoring Practice Problems - Set 2.

## Factoring Using the Associative Property

The associative and distributive properties can be confusing. The associative property states (ab) c = a (bc). This property says that when multiplying three (or more) quantities you can multiply the first two then the third or multiply the second two then the first. For example, it might be tempting to write 5(x + 1)(y – 3) = (5x + 5)(5y – 15). But (5x + 5)(5y – 15) = [5(x + 1)][5(y – 3)] = 25(x + 1)(y – 3). The “5” can be grouped either with “x + 1” or with “y – 3” but not both: [5(x + 1)] (y – 3) = (5x + 5)(y – 3) or (x + 1)[5(y – 3)] = (x + 1)(5y – 15)

Factors themselves can have more than one term. For instance 3 (x + 4) – x (x + 4) has x + 4 as a factor in each term, so x + 4 can be factored from 3(x + 4) and x (x + 4):

3(x + 4) – x(x + 4) = (3x)(x + 4).

#### Examples

2x (3x + y) + 5y (3x + y)

= (2x + 5y)(3x + y)

10 y (xy)+ xy

= 10y(xy)+ 1(xy)

= (10y + 1)(xy)

8 (2x – 1)+ 2x (2x – 1) – 3y(2x – 1)

= (8 + 2x3y )(2x – 1)

Find practice problems and solutions at Factoring Practice Problems - Set 3.

## More Factoring

An algebraic expression raised to different powers might appear in different terms. Factor out this expression raised to the lowest power.

#### Examples

6(x + 1)2 – 5(x + 1)

= [6(x + 1) ](x + 1) – 5 (x + 1)

= [6(x + 1)5 ](x + 1) = (6x + 6 – 5)(x + 1)

= (6x + 1)(x + 1)

10(2x – 3)3 + 3(2x – 3)2

= [10(2x – 3) ](2x – 3)2 + 3(2x – 3)2

= [10(2x – 3) + 3 ](2x – 3)2

= (20x – 30 + 3)(2x – 3)2

= (20x – 27)(2x – 3)2

9(14x + 5)4 + 6x(14x + 5) – (14x + 5)

= [9(14x + 5)3 ](14x + 5) + 6x(14x + 5) – 1(14x + 5)

= [9(14x + 5)3 + 6x – 1](14x + 5)

Find practice problems and solutions at Factoring Practice Problems - Set 4.

More practice problems for this concept can be found at: Algebra Factoring Practice Test.

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