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

Published by Pearson

# Chapter 7 - Section 7.2 - Multiplying and Dividing Rational Expressions - Exercise Set - Page 501: 76

#### Answer

$\Big(\dfrac{x^{2}-9}{x^{2}-1}\cdot\dfrac{x^{2}+2x+1}{2x^{2}+9x+9}\Big)\div\dfrac{2x+3}{1-x}=-\dfrac{(x-3)(x+1)}{(2x+3)^{2}}$

#### Work Step by Step

$\Big(\dfrac{x^{2}-9}{x^{2}-1}\cdot\dfrac{x^{2}+2x+1}{2x^{2}+9x+9}\Big)\div\dfrac{2x+3}{1-x}$ Factor the expressions inside the parentheses completely: $\Big(\dfrac{(x-3)(x+3)}{(x-1)(x+1)}\cdot\dfrac{(x+1)^{2}}{(x+3)(2x+3)}\Big)\div\dfrac{2x+3}{1-x}=...$ Evaluate the product of the two rational expressions inside the parentheses: $...=\dfrac{(x-3)(x+3)(x+1)^{2}}{(x-1)(x+1)(x+3)(2x+3)}\div\dfrac{2x+3}{1-x}=...$ Evaluate the division: $...=\dfrac{(x-3)(x+3)(x+1)^{2}(1-x)}{(x-1)(x+1)(x+3)(2x+3)^{2}}=...$ Simplify by removing the factors that appear both in the numerator and the denominator. To eliminate $(1-x)$ and $(x-1)$, change the sign of the first factor and also change the sign of the fraction: $...=-\dfrac{(x-3)(x+3)(x+1)^{2}(x-1)}{(x-1)(x+1)(x+3)(2x+3)^{2}}=-\dfrac{(x-3)(x+1)}{(2x+3)^{2}}$

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