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

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Answer 1 · 2017-01-17T14:21:33

The answer is #=(3-z)/(2z-5)#

We use

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

The numerator is

#9-z^2=(3-z)(3+z)#

The denominator is

#2z^2+z-15=(2z-5)(z+3)#

So,

#(9-z^2)/(2z^2+z-15)=((3-z)cancel((3+z)))/((2z-5)cancel((z+3)))#

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