Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.7 - Factoring by Special Products - Exercise Set - Page 309: 23



Work Step by Step

Using factoring techniques, then, \begin{array}{l} x^2+6x+9-y^2 \\= (x^2+6x+9)-y^2 \\= (x+3)^2-y^2 \\= [(x+3)-y][(x+3)+y] \\= (x+3-y)(x+3+y) .\end{array}
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