Elementary Algebra

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

Chapter 7 - Algebraic Fractions - Chapters 1-7 Cumulative Review Problem Set: 25

Answer

$-15n^{4}-18n^{3}+6n^{2}$

Work Step by Step

First, we multiply each term inside the parenthesis by $3n^{2}$. Then, we use the rule $a^{m}\times a^{n}=a^{m+n}$ to simplify, $(-3n^{2})(5n^{2}+6n-2)$ =$(-3n^{2})(5n^{2})+(-3n^{2})(6n)+(-3n^{2})(-2)$ =$(-3\times5\times n^{2}\times n^{2})+(-3\times6\times n^{2}\times n)+(-3\times-2\times n^{2})$ =$(-15\times n^{2+2})+(-18\times n^{2+1})+(6\times n^{2})$ =$(-15\times n^{4})+(-18\times n^{3})+(6\times n^{2})$ =$(-15n^{4})+(-18n^{3})+(6n^{2})$ =$-15n^{4}-18n^{3}+6n^{2}$
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