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 - 5.6 Factoring: A General Strategy - 5.6 Exercise Set: 15

Answer

$3t(3t+1)(3t-1)$

Work Step by Step

Factoring the $GCF= 3t $, the given expression, $ 27t^3-3t ,$ is equivalent to \begin{array}{l} 3t(9t^2-1) .\end{array} Using $a^2-b^2=(a+b)(a-b)$, the expression above is further factored as \begin{array}{l} 3t(3t+1)(3t-1) .\end{array}
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