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.8 Solving Equations - A.8 Assess Your Understanding - Page A70: 41

Answer

$-1$ (multiplicity 2) and $1$

Work Step by Step

$x^3+x^2-x-1=0$, $x^2(x+1)-(x+1)=0$, $(x+1)(x^2-1)=0$, $(x+1)(x+1)(x-1)=0$, thus $x=-1$ (multiplicity 2) and $x=1$
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