How do you simplify $\frac{9 - z^{2}}{2 z^{2} + z - 15}$ $(9-z^2)/(2z^2+z-15)$ ?

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Answer 1 · 2017-01-17T14:21:33

The answer is $= \frac{3 - z}{2 z - 5}$ $=(3-z)/(2z-5)$

We use

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

The numerator is

$9 - z^{2} = (3 - z) (3 + z)$ $9-z^2=(3-z)(3+z)$

The denominator is

$2 z^{2} + z - 15 = (2 z - 5) (z + 3)$ $2z^2+z-15=(2z-5)(z+3)$

So,

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

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

How do you simplify (9-z^2)/(2z^2+z-15)9−z22z2+z−15(9-z^2)/(2z^2+z-15)?