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 - Page 313: 47

Answer

$2(x-3)(x^2+3x+9)$

Work Step by Step

Factoring the $GCF= 2 $ results to $ 2(x^3-27) $. Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of two cubes, then, \begin{array}{l} 2(x^3-27) \\= 2[(x)+(-3)][(x)^2-(x)(-3)+(-3)^2] \\= 2(x-3)(x^2+3x+9) .\end{array}
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