Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Sections 5.1-5.7 - Integrated Review - Operations on Polynomials and Factoring Strategies: 11

Answer

$x(x-1)(x^2+x+1)$

Work Step by Step

Factoring the $GCF=x$, then the given expression, $ x^4-x $, is equivalent to \begin{array}{l} x(x^3-1) .\end{array} Using $a^3-b^3=(a-b)(a^2+ab+b^2)$ or the factoring of the difference of 2 cubes, then, \begin{array}{l}\require{cancel} x(x^3-1) \\\\= x(x-1)(x^2+x+1) .\end{array}
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