# How do you evaluate \frac { 2( x ^ { 2} + 32) } { ( x + 8) ( x - 8) } - \frac { x - 4} { x + 8}?

Feb 1, 2017

$\frac{x + 4}{x - 8}$

#### Explanation:

Take out $\frac{1}{x + 8}$ as a common factor

$\frac{1}{x + 8} \left[\frac{2 \left({x}^{2} + 32\right)}{x - 8} - \left(x - 4\right)\right]$

$\frac{1}{x + 8} \left[\frac{2 \left({x}^{2} + 32\right) - \left(x - 4\right) \left(x - 8\right)}{x - 8}\right]$

$\frac{1}{x + 8} \left[\frac{2 {x}^{2} + 64 - {x}^{2} + 12 x - 32}{x - 8}\right]$

$\frac{{x}^{2} + 12 x + 32}{\left(x + 8\right) \left(x - 8\right)}$

Now factorise the numerator

$\frac{\left(x + 8\right) \left(x + 4\right)}{\left(x + 8\right) \left(x - 8\right)}$

$\frac{x + 4}{x - 8}$

Feb 1, 2017

$= \frac{x + 4}{x - 8}$

#### Explanation:

$\frac{2 \left({x}^{2} + 32\right)}{\left(x + 8\right) \left(x - 8\right)} - \frac{x - 4}{x + 8}$

$= \frac{2 \left({x}^{2} + 32\right) - \left(x - 4\right) \left(x - 8\right)}{\left(x + 8\right) \left(x - 8\right)}$

$= \frac{2 \left({x}^{2} + 32\right) - \left({x}^{2} - 8 x - 4 x + 32\right)}{\left(x + 8\right) \left(x - 8\right)}$

$= \frac{2 {x}^{2} + 64 - {x}^{2} + 12 x - 32}{\left(x + 8\right) \left(x - 8\right)}$

$= \frac{{x}^{2} + 12 x + 32}{\left(x + 8\right) \left(x - 8\right)}$

$= \frac{{x}^{2} + 8 x + 4 x + 32}{\left(x + 8\right) \left(x - 8\right)}$

$= \frac{x \left(x + 8\right) + 4 \left(x + 8\right)}{\left(x + 8\right) \left(x - 8\right)}$

$= \frac{\left(x + 8\right) \left(x + 4\right)}{\left(x + 8\right) \left(x - 8\right)}$

$= \frac{x + 4}{x - 8}$