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: 53

Answer

zeros:$\displaystyle \qquad -\frac{1}{2},\ -1,\ \ -3\pm \sqrt{10}$

Work Step by Step

$P(x)=2x^{4}+15x^{3}+17x^{2}+3x-1,$ ($1$ sign variations, we expect $1$ positive real zero) $P(-x)=2x^{4}-15x^{3}+17x^{2}-3x-1,$ ($3$ sign variations, we expect $3$ or $1$ negative real zeros) $a_{0}=-1 \quad $p: $\pm 1,$ $a_{n}=2,\qquad $q: $\pm 1,\pm 2$ Possible rational zeros $\displaystyle \frac{p}{q}:\quad$ with synthetic division, we try$ \pm 1,\pm 1/2,$... $\begin{array}{r|rrrrrrrrrrr}\hline -1 & 2 & 15 & 17 & 3 & -1 & & \\ & & -2 & -13 & -4 & 1 & & \\ \hline & 2 & 13 & 4 & -1 & 0 & & \\ & & & & & & & \\ - 1/2 & 2 & 13 & 4 & -1 & & & \\ & & -1 & -6 & 1 & & & \\ \hline & 2 & 12 & -2 & 0 & & & \\ \end{array}$ $P(x)=(x+\displaystyle \frac{1}{2})(x+1)(2x^{2}+12x-2)=2(x+\frac{1}{2})(x+1)(x^{2}+6x-1)$ Find the zeros of the last factor with the quadratic formula $x=\displaystyle \frac{-6\pm\sqrt{(6)^{2}-4(1)(-1)}}{2(1)}$ $=\dfrac{-6\pm\sqrt{40}}{2}=\dfrac{-6\pm 2\sqrt{10}}{2}=-3\pm \sqrt{10}$ zeros:$\displaystyle \qquad -\frac{1}{2},\ -1,\ \ -3\pm \sqrt{10}$
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