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 6.5 - A General Factoring Strategy - Exercise Set - Page 462: 51



Work Step by Step

$ y^{3}+2y^{2}-4y-8\qquad$...factor in groups. $=(y^{3}+2y^{2})+(-4y-8)$ $=y^{2}(y+2)-4(y+2)\qquad$...factor out the common binomial factor,$y+2$. $=(y+2)(y^{2}-4)\qquad$...recognize the difference of two squares: $a^{2}-b^{2}=(a-b)(a+b)$ $=(y+2)(y-2)(y+2)$ $=(y+2)^{2}(y-2)$
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