# How do you simplify (2x + 10) /( x - 1) times (x^2 - 1) /( x + 5) - 4 /(x + 1)?

May 23, 2017

$= \frac{2 {\left(x + 1\right)}^{2} - 4}{x + 1}$

$= \frac{2 \left({x}^{2} + 2 x - 1\right)}{\left(x + 1\right)}$

#### Explanation:

Factorise wherever possible.

$\frac{2 x + 10}{x - 1} \times \frac{{x}^{2} - 1}{x + 5} - \frac{4}{x + 1} \text{ } \leftarrow$ there are 2 terms

$= \textcolor{b l u e}{\frac{2 \left(x + 5\right)}{\left(x - 1\right)} \times \frac{\left(x + 1\right) \left(x - 1\right)}{\left(x + 5\right)}} - \textcolor{red}{\frac{4}{\left(x + 1\right)}}$

You may cancel in a term, as long as it is through a $\times$ sign:

$= \textcolor{b l u e}{\frac{2 \cancel{\left(x + 5\right)}}{\cancel{\left(x - 1\right)}} \times \frac{\left(x + 1\right) \cancel{\left(x - 1\right)}}{\cancel{\left(x + 5\right)}}} - \textcolor{red}{\frac{4}{\left(x + 1\right)}}$

$= \textcolor{b l u e}{\frac{2 \left(x + 1\right)}{1}} - \textcolor{red}{\frac{4}{\left(x + 1\right)}} \text{ } \leftarrow$ find LCD and subtract

$= \frac{2 \left(x + 1\right) \left(x + 1\right) - 4}{\left(x + 1\right)}$

$= \frac{2 {\left(x + 1\right)}^{2} - 4}{x + 1}$

Expanding further gives:
$\frac{2 \left({x}^{2} + 2 x + 1\right) - 4}{\left(x + 1\right)}$

$= \frac{2 {x}^{2} + 4 x + 2 - 4}{\left(x + 1\right)}$

$= \frac{2 {x}^{2} + 4 x - 2}{\left(x + 1\right)}$

$= \frac{2 \left({x}^{2} + 2 x - 1\right)}{\left(x + 1\right)}$