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: 28


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

Work Step by Step

Factor what we can: $x^{2}+x=x(x+1)$ $x^{2}-4=(x-2)(x+2)\quad $(diff. of squares) $x^{2}-1=(x-1)(x+1)\quad $(diff. of squares) $ x^{2}+5x+6\quad$ ... find factors of c whose sum is b... ... we find 2 and 3 ... $=(x+2)(x+3)$ Expression $=\displaystyle \frac{x(x+1)}{(x-2)(x+2)}\div\frac{(x-1)(x+1)}{(x+2)(x+3)}$ ... Division with $\displaystyle \frac{a}{b}$ equals multiplication with $\displaystyle \frac{b}{a}.$ (neither a or b can be 0, because of the denominator we exclude $-2$ and $-3$, numerator: exclude $-1$ and $1$) $=\displaystyle \frac{x(x+1)}{(x-2)(x+2)}\cdot\frac{(x+2)(x+3)}{(x-1)(x+1)}$ ...exclude any additional values that yield 0 in the denominator: $x\neq-1, -2, -3,1,2$ ... cancel common factors $=\displaystyle \frac{x(x+3)}{(x-2)(x-1)},\qquad x\neq -1, -2, -3,1,2$
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