Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.4 - A General Approach to Factoring - 5.4 Exercises: 37

Answer

$(8+10z)(64-80z+100z^2)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 512+1000z^3 ,$ use the factoring of the sum or difference of $2$ cubes. $\bf{\text{Solution Details:}}$ The expressions $ 512 $ and $ 1000z^3 $ are both perfect cubes (the cube root is exact). Hence, $ 512+1000z^3 $ is a $\text{ sum }$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to \begin{array}{l}\require{cancel} (8)^3+(10z)^3 \\\\= (8+10z)[(8^2)-8(10z)+(10z)^2] \\\\= (8+10z)(64-80z+100z^2) .\end{array}
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