Encyclopedia>Algebra>Basic Algebra>Fractions and Factorization>How to Factorize and Simplify Fractions with Variables

Factorization and simplifying are among the most useful procedures in mathematics. If they’re not second nature to you yet, reach out to your teacher, to a friend, or call someone for help. This needs to be on point!

Theory

Factorization is rewriting a sum or difference as a product.

There are several ways to factorize an expression. One is to write a number as a product:

$$45=9\cdot 5$$ |

Another is to factorize an expression containing variables. You can do this by putting common factors outside parentheses,

$$35x+7=7\cdot 5\cdot x+7\cdot 1=7\phantom{\rule{-0.17em}{0ex}}\left(5x+1\right),$$ |

turning the expression into a product of parentheses,

$${x}^{2}-x-12=\phantom{\rule{-0.17em}{0ex}}\left(x+3\right)\phantom{\rule{-0.17em}{0ex}}\left(x-4\right),$$ |

or turning the expression into a product of parentheses and numbers/variables: $$\begin{array}{llll}\hfill 3{x}^{2}+3x-6& =3\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}+x-2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3\phantom{\rule{-0.17em}{0ex}}\left(x+2\right)\phantom{\rule{-0.17em}{0ex}}\left(x-1\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

**Note!** Remember that if you put an entire term outside the parentheses, the factor $1$ will still remain inside the parentheses.

Theory

Simplifying is accomplished by dividing common factors by each other, because $\frac{x}{x}=1$. This means that when you have the same factor in the numerator and denominator, they cancel each other out, and are replaced with $\cdot 1$!

When you multiply or divide something by 1, the value stays the same. When you cancel common factors, this is exactly what happens, because $\frac{\text{factor}}{\text{factor}}=1$.

You will often have to factorize before you can simplify. This is because expressions are often sums or differences. If you try to factorize and there are no common factors, then the expression is already as simplified as it can be.

Example 1

$$\frac{2x+2}{x+1}=\frac{2\text{(x+1)}}{\text{x+1}}=2$$ |

Example 2

$$\frac{4{x}^{3}+4x}{{x}^{2}+1}=\frac{4x\text{(x2+1)}}{\text{x2+1}}=4x$$ |

Example 3

$$\begin{array}{llll}\hfill \frac{36{x}^{2}-18x}{20x-60}& =\frac{6x\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)}{20\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{6x\text{(x\u22123)}}{20\text{(x\u22123)}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\text{6}x}{\text{20}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{3x}{10}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

Example 4

$$\begin{array}{llll}\hfill \frac{24x+6}{4{x}^{2}+5x+1}& =\frac{6\phantom{\rule{-0.17em}{0ex}}\left(4x+1\right)}{\phantom{\rule{-0.17em}{0ex}}\left(4x+1\right)\phantom{\rule{-0.17em}{0ex}}\left(x+1\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{6\text{(4x+1)}}{\text{(4x+1)}\phantom{\rule{-0.17em}{0ex}}\left(x+1\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{6}{x+1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$

Example 5

$$\begin{array}{llll}\hfill \frac{2{x}^{2}+2x-12}{{x}^{2}-x-2}& =\frac{2\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}+x-6\right)}{\phantom{\rule{-0.17em}{0ex}}\left(x+1\right)\phantom{\rule{-0.17em}{0ex}}\left(x-2\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{2\phantom{\rule{-0.17em}{0ex}}\left(x+3\right)\text{(x\u22122)}}{\phantom{\rule{-0.17em}{0ex}}\left(x+1\right)\text{(x\u22122)}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{2x+6}{x+1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$