Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.4 - Polynomials - Exercise Set - Page 56: 89



Work Step by Step

Assuming $x\ne\frac{7}{2}$, we have $\frac{(2x-7)^5}{(2x-7)^3}=(2x-7)^2=4x^2-28x+49$
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