Answer
$\frac{2-20x^3+45x}{3(4x^2-9)^4}$
Work Step by Step
Start with cancelling the common factor $5(4x^2-9)^2$, we have:
$\frac{10(4x^2-9)^2-25x(4x^2-9)^3}{15(4x^2-9)^6}=\frac{2-5x(4x^2-9)}{3(4x^2-9)^4}=\frac{2-20x^3+45x}{3(4x^2-9)^4}$
Precalculus (6th Edition)
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
Published by
Pearson
ISBN 10:
013421742X
ISBN 13:
978-0-13421-742-0
Chapter R - Review of Basic Concepts - R.6 Rational Exponents - R.6 Exercises - Page 65: 114
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