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: 19


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

Work Step by Step

Factoring $x^{2}+bx+c $ we search for two factors of c (m and n) such that m+n=b. If they exist, $x^{2}+bx+c =(x+m)(x+n)$ Factor what you can: $ x^{2}-5x+6= \quad$... we find factors $-3$ and $+4,$ $=(x-3)(x-2)$ $ x^{2}-2x-3\quad$... we find factors $-3$ and $+1,$ $=(x-3)(x+1)$ $x^{2}-1 =$ difference of squares $= (x-1)(x+1)$ $x^{2}-4 =$ difference of squares $= (x-2)(x+2)$ Expression $=\displaystyle \frac{(x-3)(x-2)}{(x-3)(x+1)}\cdot\frac{(x+1)(x-1)}{(x-2)(x+2)}$ $\qquad$ ...exclude the values that yield 0 in the denominator: $x\neq-2, -1,2,3$ ... cancel common factors: $=\displaystyle \frac{x-1}{x+2},\qquad x\neq-2, -1,2,3$
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