Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 4 - Review Exercises: 19

Answer

$-12x^{2}y^{8}$

Work Step by Step

$(-3x^{4}y^{3})(4x^{-2}y^{5})=-3\times4\times x^{4}\times x^{-2}\times y^{3}\times y^{5}=-12\times x^{4+(-2)}\times y^{3+5}=-12\times x^{2}\times y^{8}=-12x^{2}y^{8}$
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