Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Appendix A - Review - A.4 Factoring Polynomials - A.4 Assess Your Understanding - Page A40: 97

Answer

$(x-3)(x+3)(x^2+9)$

Work Step by Step

We have $ x^4-81=x^4-3^4$. The polynomial $x^4-3^4$ is a difference of two squares. That is, by using the fact that $a^2-b^2=(a-b)(a+b)$, we have $$ x^4-3^4=(x^2-3^2)(x^2+3^2)=(x-3)(x+3)(x^2+9) .$$ Moreover, the polynomial $x^2+9$ is a prime polynomial and can not be factored further.
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