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: 53


$\displaystyle \frac{x(x-1)}{(x+5)(x-2)(x+3)}, \quad x\neq-5,2, -3$

Work Step by Step

Factor each denominator. Factoring $x^{2}+bx+c$, we search for two factors of c (m and n) such that m+n=b. If they exist, $x^{2}+bx+c =(x+m)(x+n)$ $ x^{2}+3x-10=\quad$... we find factors $+5$ and $-2,$ $=(x+5)(x-2)$ $ x^{2}+x-6=\quad$... we find factors $+3$ and $-2,$ $=(x+3)(x-2)$ $LCD=(x+5)(x-2)(x+3)$ $\displaystyle \frac{3x}{x^{2}+3x-10}-\frac{2x}{x^{2}+x-6}=\frac{3x}{(x+5)(x-2)}-\frac{2x}{(x+3)(x-2)}$ $=\displaystyle \frac{3x}{(x+5)(x-2)}\times\frac{x+3}{x+3}-\frac{2x}{(x+3)(x-2)}\times\frac{x+5}{x+5}$ $=\displaystyle \frac{3x^{2}+9x}{(x+5)(x-2)(x+3)}-\frac{2x^{2}+10x}{(x+5)(x-2)(x+3)}$ $=\displaystyle \frac{3x^{2}+9x-2x^{2}-10x}{(x+5)(x-2)(x+3)}$ $=\displaystyle \frac{x^{2}-x}{(x+5)(x-2)(x+3)}$ $=\displaystyle \frac{x(x-1)}{(x+5)(x-2)(x+3)}, \quad x\neq-5,2, -3$
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