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: 61

Answer

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

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: $=2^3x^3-1^3 \\(2x)^3-1^3$ The binomial above is a difference of two cubes. Factor using the formula above with $a=2x$ and $b=1$ to obtain: $=(2x-1)[(2x)^2+2x(1) + 1^2) \\=(2x-1)(4x^2+2x+1)$
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