Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Review Exercises - Page 143: 98



Work Step by Step

This is a difference of squares, $ x^{4}=(x^{2})^{2}$ $16=4^{2}$ We apply the formula $(A-B)(A+B)=A^{2}-B^{2}$ $=(x^{2}+4)(x^{2}-4)$ The first parentheses are a sum of squares, which is prime; the second parentheses hold another difference of squares $ x^{2}-2^{2}$, = $(x^{2}+4)(x-2)(x+2)$
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