Question 196749
(x^2+7x+12)/(x-5)*(2x-10)/(x+3)    expand the numerator for (x^2+7x+12)=(x+3)(x+4)

= ((x+3)(x+4))/(x-5)*2(x-5)/(x-3)  factor a 2 from (2x-10)= 2(x-5)

now we see that (x+3) and (x-5) cancel out leaving us with = 2(x+4) = 2x+8