## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Review - Exercises - Page 134: 43

#### 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 $\dfrac{3x+9}{x+4}$

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