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: 89

Answer

$(x+4)(x-4)(y-2)$

Work Step by Step

$x^{2}y-16y+32-2x^{2}=$... factor in pairs $=(x^{2}y-16y)+(32-2x^{2})=$ $=y(x^{2}-16)+2(16-x^{2})$... factor out -1 in the 2nd term $=y(x^{2}-16)-2(x^{2}-16)$ $=(x^{2}-16)(y-2)$ ...The first parentheses hold a difference of squares $(x)^{2}-(4)^{2}$ = $(x+4)(x-4)(y-2)$
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