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)#