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: 56

Answer

a. Zeros:$ -3, -1,2$ b. see image

Work Step by Step

$P(x)= -x^{3}-2x^{2}+5x+6,$ $a_{0}=6 \quad $p: $\pm 1,\pm 2,\pm 3,\pm 6$ $a_{n}=-1,\qquad $q: $\pm 1,$ Possible rational zeros $\displaystyle \frac{p}{q}:\quad$ with synthetic division, we try $\pm 1,\pm 2,\pm 3$... $\begin{array}{r|rrrrrrrrrrr}\hline -1 & -1 & -2 & 5 & 6 & & & \\ & & 1 & 1 & -6 & & & \\ \hline & -1 & -1 & 6 & 0 & & & \\ & & \swarrow & & & & & \\ 2 & -1 & -1 & 6 & & & & \\ & & -2 & -6 & & & & \\ \hline & -1 & -3 & 0 & & & & \\ \end{array}$ $P(x)=(x+1)(x-2)(-x-3)=-(x+1)(x-2)(x+3)$ Zeros:$ -3, -1,2$ $P(0)=6\qquad $(y intercept) Calculate $P(-0.5)=3.125$ and $P(0.5)=7.875$ to figure whether the graph turns before or after the y-intercept. (Here, the graph rises through the y-intercept.) The leading coefficient is negative, so the left far end is pointing up. From left to right, the graph comes from far above x, falling, crosses x at (-3,0), dives down, turns, crosses x again at (-1,0), rises through the y-intercept at (0,6), turns, falls to the next zero at (2,0), from where it continues falling.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.