Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - Connecting: The Concepts - Exercises - Page 350: 8



Work Step by Step

Distribute each term of the first factor to the second factor to obtain: $=3a(2a-5) -2(2a-5)$ Distribute $3a$ and $2$ to obtain: $=3a(2a)-3a(5) -2(2a)-2(-5) \\=6a^2-15a-4a-(-10) \\=6a^2-19a+10$
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