Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.4 - Multiplying Polynomials - Exercise Set - Page 289: 51

Answer

$-24a^2b^{3}+40a^2b^{2}-160a^2b$

Work Step by Step

\text{ Using the Distributive Property, the given expression is equivalent to }\\ \begin{align*} & -8a^2b(3b^2-5b+20) \\\\&= -24a^2b^{1+2}+40a^2b^{1+1}-160a^2b \\\\&= -24a^2b^{3}+40a^2b^{2}-160a^2b .\end{align*}
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