Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.5 - Factoring Polynomials - Exercise Set - Page 69: 62



Work Step by Step

RECALL: A difference of two cubes can be factored using the formula: $a^3-b^3 = (a-b)(a^2+ab+b^2)$ The given expression can be written as: $=3^3x^3-1^3 \\(3x)^3-1^3$ The binomial above is a difference of two cubes. Factor using the formula above with $a=3x$ and $b=1$ to obtain: $=(3x-1)[(3x)^2+3x(1) + 1^2) \\=(3x-1)(9x^2+3x+1)$
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