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: 93



Work Step by Step

Recall that $a^2-b^2=(a+b)(a-b)$. Since $x^2+16$ is in this form, we would use this method to factor. However, as $\sqrt{-16}$ is imaginary, there are no real factors, and thus the expression is $\text{prime}$.
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