Answer
(2, 0) (-2,0) (-3,0)
See graph below.
Work Step by Step
$P(x) = x^3 + 3x^2 - 4x - 12$
$P(x) = x^2 (x + 3) - 4 (x + 3)$
$P(x) = (x^2 - 4) (x+3)$
$P(x) = (x-2) (x+2) (x+3)$
Thus the zeros are (2, 0) (-2,0) (-3,0)
Precalculus: Mathematics for Calculus, 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305071751
ISBN 13:
978-1-30507-175-9
Chapter 3 - Section 3.2 - Polynomial Functions and Their Graphs - 3.2 Exercises - Page 266: 38
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