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

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Answer 1 · 2018-05-15T10:46:18

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

#(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 !

Answer 2 · 2018-05-15T11:51:54

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)]#