How do you simplify (2z^2 - 11z + 15)/(z^2 - 9)2z211z+15z29?

2 Answers
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))

=(2(z(z-5/2)-3(z-5/2)))/((z-3)(z+3))

=(2cancel((z-3))(z-5/2))/(cancel((z-3))(z+3))

=(2(z-5/2))/(z+3)=(2z-5)/(z+3)

\0/ here's our answer !

May 15, 2018

First we need to split the middle term

After splitting... you'll get

(2z^2-6z-5z+15)/(z^2-3^2)

This is a law of exponents

a^2-b^2=(a+b)(a-b)

Factorize

(2z(z-3)-5(z-3))/((z-3)(z+3))

You get

((2z-5)cancel((z-3)))/(cancel((z-3))(z+3))

You are left with

color(red)[(2z-5)/(z+3)]