Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 9 - Roots and Radicals - Chapters 1-9 Cumulative Review Problem Set - Page 432: 17



Work Step by Step

Since all terms are multiplied by each other, we can collect constants and variables separately in their own parentheses. We then use the rule $a^{m}\times a^{n}=a^{m+n}$ and $\frac{1}{x^{2}}=x^{-2}$ to simplify the expression: $(-4x^{-5})(2x^{3})$ =$(\frac{-4}{x^{5}})(2x^{3})$ =$(-4\times2)\times(\frac{x^{3}}{x^{5}})$ =$(-8)\times x^{3-5}$ =$(-8)\times x^{-2}$ =$\frac{-8}{x^{2}}$ =$-\frac{8}{x^{2}}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.