Intermediate Algebra (6th Edition)

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

Chapter 5 - Sections 5.1-5.7 - Integrated Review - Operations on Polynomials and Factoring Strategies - Page 313: 35

Answer

$(x^2+1)(x+1)(x-1)$

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^4-1 \\= (x^2+1)(x^2-1) \\= (x^2+1)(x+1)(x-1) .\end{array}
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