Factoring a polynomial is to write it as the product of simpler polynomials.

Example:

2 *x* + 4 = 2(*x* + 2)

3 *x*^{ 2} - *x* = *x*(3*x* - 1)

NOTE: it is very easy to check if your factorization is correct by multiplying to see if you get the original polynomial

Example: check that 3 *x*^{ 2} - *x* = *x*(3*x* - 1)

Expand *x*(3*x* - 1) by mulitplication

*x*(3*x* - 1) = (*x*)(3*x*) +(*x*)(-1) = 3*x*^{2} - *x* , which is correct.

__What is factorization by common factor?__

It is a factorization method based on the law of distributivity

*a(b + c) = a · b + a · c *

used in reverse as follows

*a · b + a · c = a(b + c)*

*a* is a common factor to *a b* and *a c* is therefore factored out.

Example: Find a common factor and use the method of distributivity in reverse to factor the polynomials completely.

a) *9 x - 6*

b) *x*^{ 2} - x

c) *3 x + 12 x y*

d) *16 x*^{ 3} + 8 x^{ 2} y + 4 x y^{ 2}

e) *2 x*^{ 4}(x + 5) + x^{ 2}(x + 5)

Solution to the above examples

a) Find any common factors in the two terms of *9 x - 6* by expressing both terms *9x* and *6* in the given binomial as prime factorization. Hence

*9 x - 6 = 3 ·3 ·x - 2 ·3*

The greatest common factor is 3 and is factored out. Hence

*9 x - 6 = 3 (3 x - 2)*

b) The prime factorization of *x*^{ 2} and *x* is needed to find the greatest common factor in *x*^{ 2} - x.

*x*^{ 2} - x = x · x - x = x · x - 1 · x

The greatest common factor is *x* and is therefore factored out. Hence

*x*^{ 2} - x = = x (x - 1)

c) The prime factorizations of *3 x* and *12 x y* are needed to find the greatest common factor in *3 x + 12 x y*.

*3 x + 12 x y = 3 · x - 3 · 4 · x · y = 3 · x · 1 - 3 x · 4 · y *

The greatest common factor is *3 x*. Hence

*3 x + 12 x y = 3 x (1 + 4 y)*

d) The prime factorization of *16 x*^{ 3} , * 8 x*^{ 2} y and * 4 x y*^{ 2} are needed to find the greatest common factor in *16 x*^{ 3} + 8 x^{ 2} y + 4 x y^{ 2}.

*16 x*^{ 3} + 8 x^{ 2} y + 4 x y^{ 2}
= *2 · 2 · 2 · 2 · x · x · x + 2 · 2 · 2 · x · x · y + 2 · 2 · x · y · y*

The greatest common factor is *2 · 2 · x = 4 x*. Hence

*16 x*^{ 3} + 8 x^{ 2} y + 4 x y^{ 2} = 4 x ( 2 · 2 · x · x + 2 · x · y + y · y) = 4 x (4 x^{ 2} + 2 x y + y^{ 2})

e) We note that *x + 5* is a common factor which can be factored out as follows:

*2 x*^{ 4}(x + 5) + x^{ 2}(x + 5) = (x + 5)(2 x^{ 4} + x^{ 2})

We now find the greatest common factor of the terms *2 x*^{ 4} and *x*^{ 2} and factor it out.

*2 x*^{ 4} + x^{ 2} = 2 · x · x · x · x + x · x = x^{ 2}(2 x^{ 2} + 1)

The complete factoring of *2 x*^{ 4}(x + 5) + x^{ 2}(x + 5) is written as follows:

*2 x*^{ 4}(x + 5) + x^{ 2}(x + 5) = x^{ 2}(x + 5)(2 x^{ 2} + 1)

Use common factors to factor completely the following polynomials.

Detailed Solutions and explanations to these questions.

a) *- 3 x + 9 *

b) *28 x + 2 x *^{ 2}

c) *11 x y + 55 x *^{ 2} y

d) *20 x y + 35 x *^{ 2} y - 15 x y ^{ 2}

e) *5 y (x + 1) + 10 y *^{ 2}(x + 1) - 15 x y (x + 1)

Detailed Solutions and explanations to these questions.