Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Appendix A - Exercise Set: 62

Answer

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

Work Step by Step

$y^4 - 16 = (y^2)^2 - (4)^2 = (y^2 - 4)(y^2 + 4)$ and then $y^2 - 4 = (y)^2 - (2)^2 = (y-2)(y+2)$ Therefore $y^4 - 16 = (y^2)^2 - (4)^2 = (y^2 - 4)(y^2 + 4) = (y+2)(y-2)(y^2+4)$
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