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: 51



Work Step by Step

Factoring the $GCF= 100 $ results to $ 100(z^3+1) $. Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of two cubes, then, \begin{array}{l} 100(z^3+1) \\= 100[(z)+(1)][(z)^2-(z)(1)+(1)^2] \\= 100(z+1)(z^2-z+1) .\end{array}
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