# How do you simplify (2z^2 - 11z + 15)/(z^2 - 9)?

May 15, 2018

(2z²-11z+15)/(z²-9)=(2(z-5/2))/(z+3)=(2z-5)/(z+3)

#### Explanation:

(2z²-11z+15)/(z²-9)

=(2(z²-11/2z+15/2))/((z-3)(z+3))

=(2(z²-5/2z-3z+15/2))/((z-3)(z+3))

$= \frac{2 \left(z \left(z - \frac{5}{2}\right) - 3 \left(z - \frac{5}{2}\right)\right)}{\left(z - 3\right) \left(z + 3\right)}$

$= \frac{2 \cancel{\left(z - 3\right)} \left(z - \frac{5}{2}\right)}{\cancel{\left(z - 3\right)} \left(z + 3\right)}$

$= \frac{2 \left(z - \frac{5}{2}\right)}{z + 3} = \frac{2 z - 5}{z + 3}$

May 15, 2018

First we need to split the middle term

After splitting... you'll get

$\frac{2 {z}^{2} - 6 z - 5 z + 15}{{z}^{2} - {3}^{2}}$

This is a law of exponents

${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

Factorize

$\frac{2 z \left(z - 3\right) - 5 \left(z - 3\right)}{\left(z - 3\right) \left(z + 3\right)}$

You get

$\frac{\left(2 z - 5\right) \cancel{\left(z - 3\right)}}{\cancel{\left(z - 3\right)} \left(z + 3\right)}$

You are left with

$\textcolor{red}{\frac{2 z - 5}{z + 3}}$