## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter P - Review Exercises - Page 143: 112

#### Answer

$\frac{(x+3)^3}{(x+2)(x-2)^2}$, $x\ne\pm2$

#### Work Step by Step

Step 1. The domain requirement of the rational expression is that $x^2-4\ne0$, which gives $x\ne\pm2$ (this covers the requirements for all the denominators). Step 2. Factor the numerator and the denominator completely. We have $\frac{x^2+6x+9}{x^2-4}\cdot\frac{x+3}{x-2}=\frac{(x+3)(x+3)}{(x+2)(x-2)}\cdot\frac{x+3}{x-2}$ Step 3. Combine the common factors from the above expression. We get the result as $\frac{(x+3)^3}{(x+2)(x-2)^2}$ with $x\ne\pm2$

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