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: 11

Answer

$(x-10)(x^2+10x+100)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ x^3-1000 ,$ use the factoring of the sum or difference of $2$ cubes. $\bf{\text{Solution Details:}}$ The expressions $ x^3 $ and $ 1000 $ are both perfect cubes (the cube root is exact). Hence, $ x^3-1000 $ is a $\text{ difference }$ 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} (x)^3-(10)^3 \\\\= (x-10)[(x)^2+x(10)+(10)^2] \\\\= (x-10)(x^2+10x+100) .\end{array}
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