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

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Answer 1 · 2015-05-26T23:32:55

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

First equation:

$2y^2-18=0$
$y^2=9$
$y=+-3 => (y-3)=0$ and $(y+3)=0$

Second one:

$(-7+-sqrt(49-4(3)(-6)))/6$
$(-7+-11)/6$
$y=-3$ , the same as $y+3=0$
$y=2/3$ , the same as $3y-2=0$

Rewriting

$(cancel(y+3)(y-3))/(cancel(y+3)(3y-2))=(y-3)/(3y-2)$

Answer 2 · 2015-05-26T23:56:55

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

$f(y) = (2(y^2 - 9))/((3y - 2)(y + 3)$ = $(2(y -3)(y + 3))/((3y - 2)(y + 3))$

$= (2(y - 3))/(3y - 2)$

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