Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 5 - Section 5.5 - Factoring Special Forms - Exercise Set - Page 372: 86

Answer

$x(6-x)(36+6x+x^2)$

Work Step by Step

Factor out $x$ to obtain: $=x(216-x^3) \\=x(6^3-x^3)$ This binomial is a difference of two cubes. RECALL: $a^3-b^3=(a-b)(a^2+ab+b^2)$ Factor the difference of two cubes using the formula above with $a=6$ and $b=x$ to obtain: $=x(6-x)(6^2+6(x) + x^2) \\=x(6-x)(36+6x+x^2)$
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