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: 48

Answer

$(2w-5)(4w^2+10w+25)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 8w^3-125 ,$ use the factoring of the sum/difference of $2$ cubes. $\bf{\text{Solution Details:}}$ Using $(a\pm b)(a^2\mp ab+b^2)$ or the factoring of the sum/difference of $2$ cubes, the expression above is equivalent to \begin{array}{l}\require{cancel} (2w-5)[(2w)^2+2w(5)+(5)^2] \\\\= (2w-5)(4w^2+10w+25) .\end{array}
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