## Algebra: A Combined Approach (4th Edition)

Published by Pearson

# Chapter 7 - Section 7.2 - Multiplying and Dividing Rational Expressions - Exercise Set: 44

#### 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: $...=\dfrac{(b+3)(b-1)(b-2)(b+2)}{(b+2)^{2}(b-1)(b+4)}=\dfrac{(b+3)(b-2)}{(b+2)(b+4)}$

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