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: 25

Answer

$-(x^2-x-8)(x^2+x+8)$

Work Step by Step

Using factoring techniques, then, \begin{array}{l} x^2+16x+64-x^4 \\= (x^2+16x+64)-x^4 \\= (x+8)^2-x^4 \\= [(x+8)-x^2][(x+8)+x^2] \\= (x+8-x^2)(x+8+x^2) \\= (-x^2+x+8)(x^2+x+8) \\= -(x^2-x-8)(x^2+x+8) .\end{array}
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