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 5.6 - A General Factoring Strategy - Exercise Set - Page 378: 6

Answer

$3(2x-1)(4x^2+2x+1)$

Work Step by Step

Consider the polynomial $(24x^3-3)$ Take out common term, that is, $3$. $\implies 3(8x^3-1)$ or, $=3[(2x)^3-1^3]$ Apply difference of cubes formula. $=3(2x-1)(4x^2+2x+1)$ Hence, $(24x^3-3)=3(2x-1)(4x^2+2x+1)$
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