Answer
See below
Work Step by Step
$$f(x)=x^4-x^3-5x^2-x-6=0\\x^4+2x^3-3x^3-6x^2+x^2+2x-3x-6=0\\x^3(x+2)-3x^2(x+2)+x(x+2)-3(x+2)=0\\(x+2)(x^3-3x^2+x-3)=0\\(x+2)(x-3)(x^2+1)=0$$
The solutions are $x+2=0\rightarrow x=-2\\x-3=0 \rightarrow x=3$
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 10 Counting Methods and Probability - 10.6 Construct and Interpret Binomial Distributions - 10.6 Exercises - Mixed Review - Page 730: 64
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