Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.4 Factoring Polynomials - R.4 Exercises - Page 44: 104



Work Step by Step

Recognize $b^{2}+8b+16=(b+4)^{2}$ $b^{2}+8b+16-a^{2}=(b+4)^{2}-a^{2}$ ... and this is a difference of squares, $A^{2}-B^{2}=(A-B)(A+B)$ $=[(b+4)+a][(b+4)-a]$ $=(b+4+a)(b+4-a)$
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