How do you simplify $\frac{2 z^{2} - 11 z + 15}{z^{2} - 9}$ $(2z^2 - 11z + 15)/(z^2 - 9)$ ?

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Answer 1 · 2018-05-15T10:46:18

$\frac{2 z ² - 11 z + 15}{z ² - 9} = \frac{2 (z - \frac{5}{2})}{z + 3} = \frac{2 z - 5}{z + 3}$ $(2z²-11z+15)/(z²-9)=(2(z-5/2))/(z+3)=(2z-5)/(z+3)$

$\frac{2 z ² - 11 z + 15}{z ² - 9}$ $(2z²-11z+15)/(z²-9)$

$= \frac{2 (z ² - \frac{11}{2} z + \frac{15}{2})}{(z - 3) (z + 3)}$ $=(2(z²-11/2z+15/2))/((z-3)(z+3))$

$= \frac{2 (z ² - \frac{5}{2} z - 3 z + \frac{15}{2})}{(z - 3) (z + 3)}$ $=(2(z²-5/2z-3z+15/2))/((z-3)(z+3))$

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

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

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

\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

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

This is a law of exponents

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

Factorize

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

You get

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

You are left with

$\frac{2 z - 5}{z + 3}$ $color(red)[(2z-5)/(z+3)]$

How do you simplify 2z2−11z+15z2−9(2z^2 - 11z + 15)/(z^2 - 9)?