# How do you simplify (2y^2 - 18)/(3y^2 + 7y - 6)?

May 26, 2015

You find the roots of both quadratic and then turn them into factors, by equaling each root to zero.

First equation:

$2 {y}^{2} - 18 = 0$
${y}^{2} = 9$
$y = \pm 3 \implies \left(y - 3\right) = 0$ and $\left(y + 3\right) = 0$

Second one:

$\frac{- 7 \pm \sqrt{49 - 4 \left(3\right) \left(- 6\right)}}{6}$
$\frac{- 7 \pm 11}{6}$
$y = - 3$, the same as $y + 3 = 0$
$y = \frac{2}{3}$, the same as $3 y - 2 = 0$

Rewriting

$\frac{\cancel{y + 3} \left(y - 3\right)}{\cancel{y + 3} \left(3 y - 2\right)} = \frac{y - 3}{3 y - 2}$

May 26, 2015

First, factor (3y^2 + 7y - 6) = (3x - 2)(x + 3) =

f(y) = (2(y^2 - 9))/((3y - 2)(y + 3) = $\frac{2 \left(y - 3\right) \left(y + 3\right)}{\left(3 y - 2\right) \left(y + 3\right)}$

$= \frac{2 \left(y - 3\right)}{3 y - 2}$