Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.1 - Exponents and Scientific Notation - Exercise Set - Page 262: 77



Work Step by Step

We are given the expression $(-4x^{2}y)(3x^{4})(-2xy^{5})$. We can simplify by first grouping like terms. $(-4\times3\times-2)\times(x^{2}\times x^{4}\times x)\times(y\times y^{5})$ Next we can use the product rule, which holds that $a^{m}\times a^{n}=a^{m+n}$ (where a is a real number, and m and n are positive integers). $24\times (x^{2+4+1})\times (y^{1+5})=24x^{7}y^{6}$
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