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: 36

Answer

Zeros:$\displaystyle \quad 2,\quad \frac{1}{3},\quad- \frac{1}{4}$ $P(x)=(3x-1)(4x+1)(x-2)$

Work Step by Step

Possible rational zeros: $\displaystyle \quad \frac{p}{q}$ where$\left\{\begin{array}{lllllll} p: & \pm 1 & \pm 2 & & & & \\ q: & \pm 1 & \pm 2, & \pm 3, & \pm 4, & \pm 6 & \pm 12 \end{array}\right.$ $P(x)$ has 2 sign changes, we expect 2 or 0 positive real zeros. $P(-x)=-12x^{3}-25x^{2}-x+3$ has 1 sign change, so we expect 1 negative zero. With synthetic division, we try the integers first. 1 is not a zero,...try 2: \begin{array}{rrrrr} 2 \ \rceil & 12 & -25 & 1 & 2 \\ & & 24 & -2 & -2 \\ \hline & 12 & -1 & -1 & 0 \\ & & & & \\ 1/3 \ \rceil & 12 & -1 & -1 & \\ & & 4 & 1 & \\ \hline & 12 & 3 & 0 & \\ & & & & \\ - 1/4 \ \rceil & 12 & 3 & & \\ & & -3 & & \\ \hline & 12 & 0 & & \\ \end{array} Zeros:$\displaystyle \quad 2,\quad \frac{1}{3},\quad- \frac{1}{4}$ $P(x)=12(x-\displaystyle \frac{1}{3})(x+\frac{1}{4})(x-2)$ $=3(x-\displaystyle \frac{1}{3})\cdot 4(x+\frac{1}{4})(x-2)$ $=(3x-1)(4x+1)(x-2)$
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