Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 10 - Analytic Geometry - Chapter Review - Cumulative Review - Page 701: 2



Work Step by Step

We are given the equation: $9x^4+33x^3-71x^2-57x-10=0$ First try to find a rational solution in the set: $\pm 1, \pm 2, \pm 5, \pm 10, \pm\dfrac{1}{3},\pm \dfrac{1}{9}, \pm\dfrac{2}{3},\pm \dfrac{2}{9}, \pm\dfrac{5}{3},\pm \dfrac{5}{9},\pm\dfrac{10}{3},\pm \dfrac{10}{9}$ Using synthetic division we get that $x_1=2$ is a solution: $(x-2)(9x^3+51x^2+31x+5)=0$ We use synthetic division again for the cubic polynomial and we get that $x_2=-5$ is also a solution: $(x-2)(x+5)(9x^2+6x+1)=0$ Factor the quadratic polynomial: $(x-2)(x+5)(3x+1)^2=0$ Find all solutions: $x_1=2$ $x_2=-5$ $3x+1=0\Rightarrow x_3=-\dfrac{1}{3},x_4=-\dfrac{1}{3}$ The solution set is: $\left\{-5,-\dfrac{1}{3},2\right\}$
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.