# Factor Polynomials by Common Factor

Questions With Solutions

How to factor a polynomial using a common factor? Questions are presented along with detailed solutions and explanations are included.

## What is factorization by common factor?

It is a factorization method based on the law of distributivity which is

*a(b + c) = a · b + a · c*

and 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.

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 expanding the factored form to check if you get the original polynomial

Example: check that 3

*x*

^{ 2}-

*x*=

*x*(3

*x*- 1)

Expand

*x*(3

*x*- 1) using the law of distributivity.

*x*(3

*x*- 1) = (

*x*)(3

*x*) +(

*x*)(-1) = 3

*x*

^{2}-

*x*, which is correct.

## More Examples

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

*9 x - 6*

b)

*x*

^{ 2}- xc)

*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*and

^{ 2}*x*is needed to find the greatest common factor in

*x*.

^{ 2}- x*x*

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

^{ 2}y*4 x y*are needed to find the greatest common factor in

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

^{ 4}*x*and factor it out.

^{ 2}*2 x*

^{ 4}+ x^{ 2}= 2 · x · x · x · x + x · x = x^{ 2}(2 x^{ 2}+ 1)The complete factoring of

*2 x*is written as follows:

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

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

## Questions

Use common factors to factor completely the following polynomialsa)

*- 3 x + 9*

b)

*28 x + 2 x*

^{ 2}c)

*11 x y + 55 x*

^{ 2}yd)

*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)### Solutions to the Above Questions

a) Find any common factors in the two terms of*- 3 x + 9*by expressing both terms

*3 x*and

*9*in the given binomial as prime factorization.

*- 3 x + 9 = - 3 · x - 3 · 3*

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

*- 3x + 9 = 3 (- x + 3) = - 3 (x - 3)*

b) Write the prime factorization of each of the terms in the given polynomial

*28 x + 2 x*.

^{ 2}*28 x + 2 x*

^{ 2}= 2 · 2 · 7 · x + 2 · x · xThe greatest common factor is

*2 x*and is factored out. Hence

*28 x + 2 x*

^{ 2}= 2 x (14 + x)c) Write the prime factorization of each of the terms in the given polynomial

*11 x y + 55 x*.

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

^{ 2}y = 11 · x · y + 5 · 11 · x · x · yThe greatest common factor is

*11 x y*and is factored out. Hence

*11 x y + 55 x*

^{ 2}y = 11 x y(1 + 5 x)d) Write the prime factorization of each of the terms in the given polynomial

*20 x y + 35 x*.

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

^{ 2}y - 15 x y^{ 2}= 2 · 2 · 5 · x · y + 5 · 7 · x · x · y - 3 · 5 · x · y · yThe greatest common factor is 5 x y and is factored out. Hence

*20 x y + 35 x*

^{ 2}y - 15 x y^{ 2}= 5 x y( 4 + 7 x - 3 y)e) We start by factoring out the common factor

*(x + 1)*in the given polynomial.

*5 y (x + 1) + 10 y*

^{ 2}(x + 1) - 15 x y (x + 1) = (x + 1)(5y + 10y^{2}- 15 x y)We now factor the polynomial

*5y + 10y*using the GCF to all three terms.

^{2}- 15 x y*5 y + 10y*+ 2 · 5 · y · y - 3 · 5 · y · x = 5 · y (1 + 2 y - 3 x)

^{2}- 15 x y = 5 · yThe given polynomial may be factored as follows.

*5 y (x + 1) + 10 y*

^{ 2}(x + 1) - 15 x y (x + 1) = 5 y(x + 1)(1 + 2y - 3 x)

### More References and links

Introduction to PolynomialsFactor Polynomials

Factor Polynomials by Grouping

Factoring of Special Polynomials High School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers

Primary Maths (Grades 4 and 5) with Free Questions and Problems With Answers

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