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 84: 22


$= \displaystyle \frac{1}{x^{2}-3x+9},\qquad x\neq-3$

Work Step by Step

Factor what you can: $x^{2}+6x+9=(x+3)^{2}\qquad $... square of a sum $ x^{3}+27= \quad$...sum of cubes = $x^{3}+3^{3},$ $=(x+3)(x^{2}-3x+9)$ ... the second numerator and denominator are factored. Expression $=\displaystyle \frac{(x+3)(x+3)}{(x+3)(x^{2}-3x+9)}\cdot\frac{1}{(x+3)}\qquad$ ...exclude the values that yield 0 in the denominator: $x\neq-3$ ... cancel common factors: $x$, and $(x-3)$ $= \displaystyle \frac{1}{x^{2}-3x+9},\qquad x\neq-3$
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