Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.4 Factoring Polynomials - R.4 Exercises - Page 43: 28


Kurt's answer is correct. The factored form of the given polynomial is $(8a-3)(2a-5)$

Work Step by Step

Factor the polynomial by grouping: Regroup: $=(16a^2-40a)+(-6a+15)$ Factor out $8a$ in the first and $-3$ in the second group to obtain: $=8a(2a-5)+(-3)(2a-5)$ Factor out $2a-5$ to obtain: $=(2a-5)[8a+(-3)] \\=(2a-5)(8a-3) \\=(8a-3)(2a-5)$ Thus, Kurt's answer is correct.
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