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: 70



Work Step by Step

$ 12y^{3}+16y^{2}-3y-4\qquad$...factor in groups. $=(12y^{3}+16y^{2})+(-3y-4)$ $=4y^{2}(3y+4)-(3y+4)\qquad$...factor out the common binomial factor,$3y+4$. $=(3y+4)(4y^{2}-1)\qquad$...recognize the difference of two squares: $a^{2}-b^{2}=(a-b)(a+b)$ $=(3y+4)(2y-1)(2y+1)$
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