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:

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

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

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

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