Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.5 Factoring Sums or Differences of Cubes - 5.5 Exercise Set: 40

Answer

$3z(z-1)(z^2+z+1)$

Work Step by Step

Factoring the $GCF= 3z^2 ,$ the given expression, $ 3z^5-3z^2 ,$ is equivalent to \begin{array}{l} 3z^2(z^3-1) .\end{array} Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, the factored form of the expression, $ 3z^2(z^3-1) ,$ is \begin{array}{l} 3z(z-1)[ (z)^2-(z)(-1)+(-1)^2] \\\\= 3z(z-1)(z^2+z+1) .\end{array}
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