Answer
$ \frac{-(x+3)(3x-1)}{(x^2+1)^2}$
Work Step by Step
$\frac{(x^2+1)\cdot3-(3x+4)\cdot2x}{(x^2+1)^2}=\frac{3x^2+3-6x^2-8x}{(x^2+1)^2}=\frac{-3x^2-8x+3}{(x^2+1)^2}=\frac{-(x+3)(3x-1)}{(x^2+1)^2}$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Appendix A - Review - A.6 Rational Expressions - A.6 Assess Your Understanding - Page A54: 91
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