Precalculus (6th Edition) Blitzer

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

Chapter P - Section P.5 - Factoring Polynomials - Exercise Set - Page 70: 134


False. The correct statement is: $x^4-16$ is factored completely as $(x^2+4)(x-2)(x+2)$.

Work Step by Step

The statement is false. $x^2-4=x^2-2^2$, a difference of two squares. A difference of two squares can still be factored using the formula: $a^2-b^2=(a-b)(a+b)$ Factor the difference of two squares using the formula above with $a=x$ and $b=2$ to obtain: $x^2-2^2=(x-2)(x+2)$ Thus, the completely factored form of $x^4-16$ is: $=(x^2+4)(x-2)(x+2)$
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