How do you simplify #(3t^2 - t - 2)/(6t^2 - 5t - 6) * (4t^2 - 9)/(2t^2 + 5t + 3)#?
1 Answer
Dec 17, 2015
Factor all the quadratics and cancel the matching terms in the numerator and denominator to find:
#(3t^2-t-2)/(6t^2-5t-6)*(4t^2-9)/(2t^2+5t+3)=(t-1)/(t+1)#
Explanation:
First factor the quadratics:
#3t^2-t-2 = (3t^2-3t)+(2t-2) = (3t+2)(t-1)#
#4t^2-9 = (2t)^2-3^2 = (2t-3)(2t+3)#
#6t^2-5t-6 = (2)(3)t^2-(3^2-2^2)t-(2)(3) = (3t+2)(2t-3)#
#2t^2+5t+3 = (2t^2 + 2t)+(3t+3) = (t+1)(2t+3)#
So:
#(3t^2-t-2)/(6t^2-5t-6)*(4t^2-9)/(2t^2+5t+3)#
#=(color(red)(cancel(color(black)((3t+2))))(t-1)color(red)(cancel(color(black)((2t-3))))color(red)(cancel(color(black)((2t+3)))))/(color(red)(cancel(color(black)((3t+2))))color(red)(cancel(color(black)((2t-3))))(t+1)color(red)(cancel(color(black)((2t+3))))#
#=(t-1)/(t+1)#
