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: 87

Answer

$(x+y)(3b+4)(3b-4)$

Work Step by Step

$9b^{2}x-16y-16x+9b^{2}y=$ Group the terms containing $b^{2}$ and factor in pairs $=(9b^{2}x+9b^{2}y)+(-16x-16y)$ $=9b^{2}(x+y)-16(x+y)$ $=(x+y)(9b^{2}-16)$ The second parentheses hold a difference of squares $(3b)^{2}-4^{2}$ = $(x+y)(3b+4)(3b-4)$
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