Answer
$\dfrac{b^{2}+2b-3}{b^{2}+b-2}\cdot\dfrac{b^{2}-4}{b^{2}+6b+8}=\dfrac{(b+3)(b-2)}{(b+2)(b+4)}$
Work Step by Step
$\dfrac{b^{2}+2b-3}{b^{2}+b-2}\cdot\dfrac{b^{2}-4}{b^{2}+6b+8}$
Factor both rational expressions completely:
$\dfrac{b^{2}+2b-3}{b^{2}+b-2}\cdot\dfrac{b^{2}-4}{b^{2}+6b+8}=\dfrac{(b+3)(b-1)}{(b+2)(b-1)}\cdot\dfrac{(b-2)(b+2)}{(b+4)(b+2)}=...$
Evaluate the product of the two rational expressions and simplify by removing the factors that appear both in the numerator and the denominator of the resulting expression:
$...=\dfrac{(b+3)(b-1)(b-2)(b+2)}{(b+2)^{2}(b-1)(b+4)}=\dfrac{(b+3)(b-2)}{(b+2)(b+4)}$
