How do you simplify #(x^2-16)/(3x^2+10x-8) #?

Question

1945 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-04T17:21:08

#(x-4)/(3x-2)#.

The Expression#=(x^2-16)/(3x^2+10x-8)#

#={(x-4)(x+4)}/{ul(3x^2+12x)-ul(2x-8)}#

#={(x-4)(x+4)}/{3x(x+4)-2(x+4)}#

#={(x-4)cancel((x+4))}/{(3x-2)cancel((x+4))}#

#=(x-4)/(3x-2)#.

Answer 2 · 2016-09-04T17:27:24

#((x-4))/((3x-2))#

In an algebraic fraction, only factors can be cancelled.

Find the factors in the numerator and denominator.

#(x^2 -16)/(3x^2 +10x -8) " " (larr"difference of squares")/(larr"quadratic trinomial")#

=#(cancel(x+4)(x-4))/((3x-2)cancel(x+4)) = ((x-4))/((3x-2))#