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: 45

Answer

zeros: $-1, \displaystyle \ \ \frac{-1\pm\sqrt{13}}{3}$

Work Step by Step

$P(x)=3x^{3}+5x^{2}-2x-4,$ ($1$ sign variations, we expect $1$ real zero) $P(-x)=-3x^{3}+5x^{2}+2x-4,$ ($3$ sign variations, we expect $3$ or $1$ negative real zeros) $a_{0}=-4 \quad $p: $\pm 1,\pm 2,\pm 4$ $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,... $\begin{array}{r|rrrrrrrrrrr}\hline -1 & 3 & 5 & -2 & -4 & & & \\ & & -3 & -2 & 4 & & & \\ \hline & 3 & 2 & -4 & 0 & & & \\ & & & & & & & \\ \end{array}$ $P(x)=(x+1)(3x^{2}+2x-4)$ Find the zeros of the second factor with the quadratic formula $x=\displaystyle \frac{-2\pm\sqrt{2^{2}-4(3)(-4)}}{2(3)}=\frac{-2\pm\sqrt{52}}{6}=\frac{-2\pm 2\sqrt{13}}{6}=\frac{-1\pm\sqrt{13}}{3}$ zeros: $-1, \displaystyle \ \ \frac{-1\pm\sqrt{13}}{3}$
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