Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.5 - Factoring Polynomials - Exercise Set - Page 69: 90

Answer

$(2x+3)(2x-3)(3y-1)$

Work Step by Step

$12x^{2}y-27y-4x^{2}+9=$... factor in pairs $=(12x^{2}y-27y)+(9-4x^{2})$... factor out -1 in the 2nd term $=3y(4x^{2}-9)-1(4x^{2}-9)$ $=(4x^{2}-9)(3y-1)$ ...The first parentheses hold a difference of squares $(2x)^{2}-(3)^{2}$ $=(2x+3)(2x-3)(3y-1)$
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