## Elementary Algebra

Published by Cengage Learning

# Chapter 7 - Algebraic Fractions - Chapter 7 Review Problem Set - Page 322: 4

#### Answer

$=\frac{4a+3}{5a-2}$

#### Work Step by Step

1. Factor the numerator $= \frac{16a^{2}+24a+9}{20a^{2}+7a-6}$ $=\frac{16a^{2}+12a+12a+9}{20a^{2}+7a-6}$ $=\frac{4a(4a+3)+3(4a+3)}{20a^{2}+7a-6}$ $=\frac{(4a+3)(4a+3)}{20a^{2}+7a-6}$ 2. Factor the denominator $=\frac{(4a+3)(4a+3)}{20a^{2}+15a-8a-6}$ $=\frac{(4a+3)(4a+3)}{5a(4a+3)-2(4a+3)}$ $=\frac{(4a+3)(4a+3)}{(5a-2)(4a+3)}$ 3. Cancel out $(4a+3)$ from the numerator and denominator: $=\frac{(4a+3)}{(5a-2)}$

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