Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 3 - Section 3.4 - Real Zeros of Polynomials - 3.4 Exercises - Page 284: 61

Answer

(a) $x=-1$ and $x=2$ (b) see graph.

Work Step by Step

(a) Possible rational zeros are $\pm1,\pm2,\pm4$, use synthetic division to find three zeros as shown in the figure $x=-1,+2,+2$ so that $P(x)=(x+1)(x-2)^2(x^2+2x+1)=(x+1)^3(x-2)^2$, thus the zeros are $x=-1$ (multiplicity 3) and $x=2$ (multiplicity 2) (b) see graph.
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