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: 44

Answer

$-6y(3y-1)(3y+1)$

Work Step by Step

Factor out $-6y$ to obtain: $=-6y(9y^2-1)$ Write $9y^2$ as $(3y)^2$ to obtain: $=-6y[(3y)^2-1]$ The binomial is a difference of two squares. RECALL: A difference of two squares can be factored using the formula: $a^2-b^2=(a-b)(a+b)$ Factor the difference of two squares using the formula above with $a=3y$ and $b=1$ to obtain: $=-6y(3y-1)(3y+1)$
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