College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Chapter P Review - Exercises - Page 77: 82


$\displaystyle \frac{x^{2}-2x-5}{(x-2)(x+1)(x+2)}$

Work Step by Step

We factor the bottom terms and create a common denominator: $\displaystyle \frac{1}{x+2}+\frac{1}{x^{2}-4}-\frac{2}{x^{2}-x-2}=\frac{1}{x+2}+\frac{1}{(x-2)(x+2)}-\frac{2}{(x-2)(x+1)} =\frac{(x-2)(x+1)}{(x-2)(x+1)(x+2)}+\frac{x+1}{(x-2)(x+1)(x+2)}-\frac{2(x+2)}{(x-2)(x+1)(x+2)}$ We distribute and combine the top terms: $=\displaystyle \frac{x^{2}-x-2+x+1-2x-4}{(x-2)(x+1)(x+2)}=\frac{x^{2}-2x-5}{(x-2)(x+1)(x+2)}$
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