Answer
$\dfrac{x^{2}+2x-3}{x^{2}+8x+16}\cdot\dfrac{3x+12}{x-1}=\dfrac{3x+9}{x+4}$
Work Step by Step
$\dfrac{x^{2}+2x-3}{x^{2}+8x+16}\cdot\dfrac{3x+12}{x-1}$
Factor both rational expressions completely:
$\dfrac{x^{2}+2x-3}{x^{2}+8x+16}\cdot\dfrac{3x+12}{x-1}=\dfrac{(x+3)(x-1)}{(x+4)^{2}}\cdot\dfrac{3(x+4)}{x-1}=...$
Evaluate the product of the two fractions and simplify by removing the factors that appear both in the numerator and in the denominator of the resulting expression:
$...=\dfrac{3(x+3)(x-1)(x+4)}{(x+4)^{2}(x-1)}=...$
$...=\dfrac{3(x+3)}{x+4}$ or $\dfrac{3x+9}{x+4}$
