Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.3 - Special Factoring - 5.3 Exercises: 55

Answer

$3(2n+3p)(4n^2-6np+9p^2)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 24n^3+81p^3 ,$ factor first the $GCF$. Then use the factoring of the sum/difference of $2$ cubes. $\bf{\text{Solution Details:}}$ Factoring the $GCF= 3 ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 3(8n^3+27p^3) .\end{array} Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or $a^3-b^3=(a-b)(a^2+ab+b^2)$ or the factoring of the sum/difference of $2$ cubes, the expression above is equivalent to \begin{array}{l}\require{cancel} 3(2n+3p)[(2n)^2-2n(3p)+(3p)^2] \\\\= 3(2n+3p)(4n^2-6np+9p^2) .\end{array}
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