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: 40

Answer

$P(x)=(x+1)(x-2)(2x-1)(3x+1)$ zeros: $-1, -1/3, 1/2, 2$

Work Step by Step

$P(x)=6x^{4}-7x^{3}-12x^{2}+3x+2,$ ($2$ sign variations, we expect $2$ or $0$ real zeros) $P(-x)=6x^{4}+7x^{3}-12x^{2}-3x+2,$ (2 sign variations, we expect 2 or 0 negative real zero) $a_{0}=2, \quad $p: $\pm 1,\pm 2,$ $a_{n}=6,\qquad $q: $\pm 1.,\pm 2,\pm 3,\pm 6$ Possible rational zeros $\displaystyle \frac{p}{q}:\quad$ with synthetic division, we try$ \pm$1, $\pm$2,... $\begin{array}{r|rrrrrrrrrrr}\hline -1 & 6 & -7 & -12 & 3 & 2 & & \\ & & -6 & 13 & -1 & -2 & & \\ \hline & 6 & -13 & 1 & 2 & 0 & & \\ & & & & & & P(x) &=(x+1)(6x^{3}-13x^{2}+x+2) \\\\ 2 & 6 & -13 & 1 & 2 & & & \\ & & 12 & -2 & -2 & & & \\ \hline & 6 & -1 & -1 & 0 & & & \\ & & & & & & P(x) &=(x+1)(x-2)(6x^{2}-x-1) \\\\ 1/2 & 6 & -1 & -1 & & & & \\ & & 3 & 1 & & & & \\ \hline & 6 & 2 & 0 & & & & \\ & & & & & & P(x) &=(x+1)(x-2)(x-\displaystyle \frac{1}{2})(6x-2) \\\\ - 1/3 & 6 & 2 & & & & & \\ & & -2 & & & & & \\ \hline & 6 & 0 & & & & & \end{array}$ $ P(x) =(x+1)(x-2)(x-\displaystyle \frac{1}{2})(x+\frac{1}{3})\cdot 6$ $P(x)= (x+1)(x-2)\displaystyle \cdot 2(x-\frac{1}{2})\cdot 3(x+\frac{1}{3})$ $P(x)=(x+1)(x-2)(2x-1)(3x+1)$ zeros: $-1, -1/3, 1/2, 2$
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