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: 334

Answer

Zeros:$\displaystyle \quad\frac{3}{2},\quad\pm 1$ $P(x)=(2x-3)(x+1)(x-1)$

Work Step by Step

Factor by grouping: $P(x)=x^{2}(2x-3)-1(2x-3)$ $=(2x-3)(x^{2}-1)$ ...the second parentheses have a difference of squares, $P(x)=(2x-3)(x+1)(x-1)$ and the zeros are $\displaystyle \quad\frac{3}{2},\quad\pm 1$
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