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 463: 114



Work Step by Step

$ 12x^{2}(x-1)-4x(x-1)-5(x-1)\qquad$...factor out the common term, $(x-1)$ $=(x-1)(12x^{2}-4x-5)$ ... Searching for two factors of $ac=-60$ whose sum is $b=-4,$ we find$\qquad 6$ and $-10.$ Rewrite the middle term and factor in pairs: $=(x-1)(12x^{2}+6x-10x-5)=$ $=(x-1)[6x(2x+1)-5(2x+1)]$ =$(x-1)(2x+1)(6x-5)$
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