Precalculus: Mathematics for Calculus, 7th Edition

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: 37

Answer

(-1, 0) and (1,0) See Graph Below

Work Step by Step

$P(x) = x^3 + x^2 - x - 1$ $P(x) = x^2 (x + 1) - 1 (x + 1)$ $P(x) = (x^2 - 1) (x+1)$ $P(x) = (x-1) (x+1) (x+1)$ Thus the zeros are (-1, 0) and (1,0)
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