Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 6 - Section 6.4 - Factoring Special Forms - Exercise Set - Page 454: 83

Answer

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

Work Step by Step

The given binomial can be written as: $(2y)^3-1^3$ The binomial above is a difference of two cubes. RECALL: A difference of two cubes can be factored using the formula: $a^3-b^3=(a-b)(a^2+ab+b^2)$ Use the formula above to factor the sum of two cubes with $a=2y$ and $b=1$ to obtain: $=(2y-1)[(2y)^2+2y(1) + 1^2] \\=(2y-1)(4y^2+2y+1)$
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