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 310: 96

Answer

$(x^{2n}+25)(x^{n}+5)(x^{n}-5)$

Work Step by Step

Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of 2 squares, then, \begin{array}{l} x^{4n}-625 \\= (x^{2n}+25)(x^{2n}-25) \\= (x^{2n}+25)(x^{n}+5)(x^{n}-5) .\end{array}
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