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: 55

Answer

a. zeros: $-2,2,3$ b. see image

Work Step by Step

$P(x)= x^{3}-3x^{2}-4x+12,$ ($4$ sign variations, we expect $4,2$ or $0$ positive real zeros) $P(-x)=-4x^{5}-18x^{4}+6x^{3}+91x^{2}+60x+9,$ ($1$ sign variations, we expect $1$ negative real zeros) $a_{0}=12 \quad $p: $\pm 1,\pm2,\pm 3,\pm 4 \pm6,\pm12$ $a_{n}=1,\qquad $q: $\pm 1$ Possible rational zeros $\displaystyle \frac{p}{q}:\quad$ with synthetic division, we try$ \pm 1,\pm 2,\pm 3$... $\begin{array}{r|rrrrrrrrrrr}\hline 2 & 1 & -3 & -4 & 12 & & & \\ & & 2 & -2 & -12 & & & \\ \hline & 1 & -1 & -6 & 0 & & & \\ & & \swarrow & & & & & \\ -2 & 1 & -1 & -6 & & & & \\ & & -2 & 6 & & & & \\ \hline & 1 & -3 & 0 & & & & \\ \end{array}$ $P(x)=(x-2)(x+2)(x-3)$ $P(0)=12\qquad $(y intercept) Calculate $P(-0.5)=13.25$ and $P(0.5)=9.375 $ to figure whether the graph turns before or after the y-intercept. (Here, the graph falls through the y-intercept.) The leading coefficient is positive, so the left far end is pointing down.\\\\ From left to right, the graph comes from far below x, crosses x at (-2,0), rises up, turns, crosses y at (0,12), falls to the next zero at (2,0), falls below x, turns, and crosses x again at (3,0), from where it continues rising.
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