Answer
(a) -2, 2, and 3
(b) See graph
Work Step by Step
(a)
$P(x)=0$
$x^3-3x^2-4x+12=0$
$(x+2)(x-2)(x-3)=0$
$x+2=0\vee x-2=0\vee x-3=0$
$x=-2\vee x=2\vee x=3$
The real zeros are $-2,2,$ and $3$.
(b) The graph of $P$ is below.
College Algebra 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305115546
ISBN 13:
978-1-30511-554-5
Chapter 3, Polynomial and Rational Functions - Section 3.4 - Real Zeros of Polynomials - 3.4 Exercises - Page 320: 55
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