How do you simplify #(12x^2+6x+90)/(6x^2-54)#?

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Answer 1 · 2016-11-10T12:41:50

#=((2x^2 +x +15))/((x+3)(x-3)#

With algebraic fractions, always try to find factors first..

#(6(2x^2 +x +15))/(6(x^2 -9))" "larr# common factor of 6

#=(6(2x^2 +x +15))/(6(x+3)(x-3)#

#=((2x^2 +x +15))/((x+3)(x-3)#

This does not simplify further.
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However, if the numerator had been #12x^2+6x-90#

We would have:

#(6(2x^2 +x -15))/(6(x^2 -9)#

#=(6(2x-5)(x+3))/(6(x+3)(x-3))#

#= ((2x+5))/((x-3))#