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: 29

Answer

$(2s+3)(3t-5)$

Work Step by Step

Regroup: $=(6st+9t)+(-10s-15)$ Factor out $3t$ in the first and $-5$ in the second group to obtain: $=3t(2s+3)+(-5)(2s+3)$ Factor out $2s+3$ to obtain: $=(2s+3)[3t+(-5)] \\=\color{blue}{(2s+3)(3t-5)}$
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