Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 3 - Review - Exercises - Page 321: 60

Answer

$-3, -1, 3/2$

Work Step by Step

This question asks for ALL zeros of the function Given $P(x) = 2x^3 + 5x^2 - 6x - 9$ See image below for synthetic division So x = -1 is a zero Factor and set P(x) = 0 $2x^2 + 3x - 9$ $(2x - 3) (x +3)$ $x = 3/2, -3$ Thus the solutions are $-3, -1, 3/2$
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