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: 52

Answer

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

Work Step by Step

$P(x)=3x^{3}-5x^{2}-8x-2,$ ($1$ sign variations, we expect $1$ positive real zero) $P(-x)=-3x^{3}-5x^{2}+8x-2,$ ($2$ sign variations, we expect $2$ or $0$ negative real zeros) $a_{0}=-2 \quad $p: $\pm 1,\pm 2$ $a_{n}=3,\qquad $q: $\pm 1,\pm 3$ Possible rational zeros $\displaystyle \frac{p}{q}:\quad$ with synthetic division, we try$ \pm 1,\pm 2, \pm 1/3$... $\begin{array}{r|rrrrrrrrrrr}\hline - 1/3 & 3 & -5 & -8 & -2 & & & \\ & & -1 & 2 & 2 & & & \\ \hline & 3 & -6 & -6 & 0 & & & \\ \end{array}$ $P(x)=(x+\displaystyle \frac{1}{3})(3x^{2}-6x-6)=3(x+\frac{1}{3})(x^{2}-2x-2)$ Find the zeros of the last factor with the quadratic formula $x=\displaystyle \frac{2\pm\sqrt{(-2)^{2}-4(1)(-2)}}{2(1)}=\frac{2\pm\sqrt{12}}{2}=\frac{2\pm 2\sqrt{3}}{2}=1\pm\sqrt{3}$ zeros:$\displaystyle \qquad -\frac{1}{3},\ \ \ 1\pm\sqrt{3}$
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