# How do you simplify \frac { x ^ { 2} - 2x - 15} { x ^ { 2} - 9} \cdot \frac { x + 3} { x - 5}?

Oct 20, 2016

$\frac{x + 3}{x - 3}$

#### Explanation:

$\setminus \frac{{x}^{2} - 2 x - 15}{{x}^{2} - 9} \setminus \cdot \setminus \frac{x + 3}{x - 5}$

Factor the fraction on the left

What are the factors of -15 that add up to -2

$\left(x - 5\right) \left(x + 3\right)$

Difference of perfect squares

$\left({x}^{2} - 9\right) = \left({x}^{2} - {3}^{2}\right) = \left(x + 3\right) \left(x - 3\right)$

Rewrite the expression using the factors you just found

$\frac{\left(x - 5\right) \left(x + 3\right)}{\left(x + 3\right) \left(x - 3\right)} \cdot \frac{x + 3}{x - 5}$

Now cross cancel

$\frac{\cancel{x - 5} \cancel{x + 3}}{\cancel{x + 3} \left(x - 3\right)} \cdot \frac{x + 3}{\cancel{x - 5}}$

You are then left with

$\frac{x + 3}{x - 3}$

Oct 20, 2016

The expression can be simplified to $\frac{x + 3}{x - 3}$.

#### Explanation:

$= \frac{\left(x - 5\right) \left(x + 3\right)}{\left(x + 3\right) \left(x - 3\right)} \times \frac{x + 3}{x - 5}$

$= \frac{x + 3}{x - 3}$

Hopefully this helps!