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: 30

Answer

Zeros: $-1,4,-1/2,4/3.$ $P(x)=(3x-4)(2x+1)(x-4)(x+1)$

Work Step by Step

Possible rational zeros: $\displaystyle \quad \frac{p}{q}$ where$\left\{\begin{array}{llllll} p: & \pm 1 & \pm 2, & \pm 4, & \pm 8, & \pm 16\\ q: & \pm 1 & \pm 2, & \pm 3, & \pm 6 & \end{array}\right.$ $P(x)$ has two sign changes, we expect 2 or 0 positive real zeros. With synthetic division, we try the integers first. 1 is not a zero ... try $-1$: \begin{array}{rrrrrr} -1 \rceil & 6 & -23 & -13 & 32 & 16 \\ & & -6 & 29 & -16 & -16 \\ \hline & 6 & -29 & 16 & 16 & 0 \\ & & & & & \\ 4 \rceil & 6 & -29 & 16 & 16 & \\ & & 24 & -20 & -16 & \\ \hline & 6 & -5 & -4 & 0 & \\ & & & & & \\ - 1/2 \rceil & 6 & -5 & -4 & & \\ & & -3 & 4 & & \\ \hline & 6 & -8 & 0 & & \\ & & & & & \\ 4/3 \rceil & 6 & -8 & & & \\ & & 8 & & & \\ \hline & 6 & -0 & & & \\ \end{array} Zeros: $-1,4,-1/2,4/3.$ $P(x)=6(x-\displaystyle \frac{4}{3})(x+\frac{1}{2})(x-4)(x+1)$ $= 3(x-\displaystyle \frac{4}{3})\cdot 2(x+\frac{1}{2})(x-4)(x+1)$ $=(3x-4)(2x+1)(x-4)(x+1)$
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.