# Factorization of Expressions

In the Math Essential Parentheses, you learned that

 $2\left(1+2\right)=2+4$

and that

 $x\left(2-3\right)=2x-3x$

But how do you simplify an expression like $2+4$ into $2\left(1+2\right)$? Let’s look at calculations in the reverse order of what we’ve been reviewing up to now.

## Parentheses and Expressions

If you’ve got an expression you’d like to factorize, you can just use the following method:

Rule

### Factorization

1.
Factorize all numbers in the expression into prime numbers.
2.
If all the numbers have a common factor (or several common factors), put parentheses around the expression and put the common factor(s) on the outside.
3.
All of the terms inside the parentheses should be divided by this factor.

You’ll be able to best learn this process through an example:

Example 1

### Only Numbers

Factorize the expression $2+4$

If you follow the method above, you have $2=2×1$ and $4=2×2$. Therefore

 $2+4=2×1+2×2$

Here 2 is a common factor that you can put outside of the parentheses, and you therefore have to remove a 2 from each term inside the parentheses:

$\begin{array}{llll}\hfill 2×1+2×2& =2\left(\text{2}×1+\text{2}×2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2\left(1+2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $2×1+2×2=2\left(\text{2}×1+\text{2}×2\right)=2\left(1+2\right)$

You should check that the two expressions give the same answer. The left-hand side is $2+4=6$, and the right-hand side is $2\left(1+2\right)=2×3=6$. Since these are equal, you know that the factorization is correct.

Factorization has a surprising number of applications, and here you’ll look at a few of them.

Often, you’ll have to put several factors outside the parentheses. Here’s an example:

Example 2

### Algebraic Expressions

Factorize the expression $3x-6x+12x$

Factorizing each term gives $\begin{array}{llll}\hfill 3x& =3\cdot 1\cdot x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 6x& =3\cdot 2\cdot x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 12x& =3\cdot 2\cdot 2\cdot x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The common factors for all terms are 3 and $x$, meaning that you get

$\begin{array}{llll}\hfill & \phantom{=}3x-6x+12x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3\cdot x-3\cdot 2\cdot x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+3\cdot 2\cdot 2\cdot x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3\cdot x\left(1-2+2\cdot 2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3x\left(1-2+4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill 3x-6x+12x& =3\cdot x-3\cdot 2\cdot x+3\cdot 2\cdot 2\cdot x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3\cdot x\left(1-2+2\cdot 2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3x\left(1-2+4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

It’s easy to check that $3x\left(1-2+4\right)=3x-6x+12x$, which tells you that you’ve gotten the right answer (try for yourself).

This gives you the following rule:

Rule

### Factorization

$\begin{array}{llll}\hfill ab+ac& =a\left(b+c\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill ab-ac& =a\left(b-c\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Here is an example with powers:

Example 3

Factorize $12{a}^{2}+3{a}^{3}+6{a}^{5}+24a$

By factoring each term separately, you see that 3 and $a$ are common factors for all terms (try yourself).

Therefore the expression becomes

$\begin{array}{llll}\hfill & \phantom{=}12{a}^{2}+3{a}^{3}+6{a}^{5}+24a\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3a\phantom{\rule{-0.17em}{0ex}}\left(4a+{a}^{2}+2{a}^{4}+8\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $12{a}^{2}+3{a}^{3}+6{a}^{5}+24a=3a\phantom{\rule{-0.17em}{0ex}}\left(4a+{a}^{2}+2{a}^{4}+8\right)$