Answer
$x=\pm\sqrt3$ or $x=2$ or $x=-2$
Work Step by Step
$x^6-4x^4-9x^2+36=(x-2)(x+2)(x^2-3)(x^2+3)=0$
Thus $x^2+3=0$ (but this is impossible since $x^2$ is non-negative) or $x^2-3=0\\x^2=3\\x=\pm\sqrt3$ or $x-2=0\\x=2$ or $x+2=0\\x=-2$
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
Published by
McDougal Littell
ISBN 10:
0618595414
ISBN 13:
978-0-61859-541-9
Chapter 5 Polynomials and Polynomial Functions - 5.4 Factor and Solve Polynomial Equations - 5.4 Exercises - Skill Practice - Page 357: 39
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