College Algebra 7th Edition

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

Chapter P, Prerequisites - Section P.7 - Rational Expressions - P.7 Exercises: 57

Answer

$\frac{-5}{(x-3)(x+2)(x+1)}$

Work Step by Step

We form a common denominator, then add, factor, and simplify: $\displaystyle \frac{1}{x^{2}+3x+2}-\frac{1}{x^{2}-2x-3}=\frac{1}{(x+2)(x+1)}-\frac{1}{(x-3)(x+1)} =\frac{x-3}{(x-3)(x+2)(x+1)}+\frac{-(x+2)}{(x-3)(x+2)(x+1)}=\frac{x-3-x-2}{(x-3)(x+2)(x+1)}=\frac{-5}{(x-3)(x+2)(x+1)}$
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