Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.6 - Rational Expressions - Exercise Set - Page 85: 51


$\displaystyle \frac{4(x+4)}{(x+3)^{2}},\quad x\neq-3$

Work Step by Step

Factor each denominator. Recognize a perfect square in the first. $\displaystyle \frac{4}{x^{2}+6x+9}+\frac{4}{x+3}=\frac{4}{(x+3)^{2}}+\frac{4}{x+3}=...$ The denominators are different. Find a common denominator. LCD=$(x+3)^{2}$ $=\displaystyle \frac{4}{(x+3)^{2}}+\frac{4}{x+3}\times\frac{x+3}{x+3}$ $=\displaystyle \frac{4}{(x+3)^{2}}+\frac{4x+12}{(x+3)^{2}}$ $=\displaystyle \frac{4+4x+12}{(x+3)^{2}}$ $=\displaystyle \frac{4x+16}{(x+3)^{2}}$ = $\displaystyle \frac{4(x+4)}{(x+3)^{2}},\quad x\neq-3$
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