Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Mid-Chapter Check Point - Page 381: 24

Answer

The roots of the polynomial equation are $x\ =\ 10$ , $x\ =\ -10$ , and $x\ =\ -\frac{5}{2}$.

Work Step by Step

$\begin{align} & 2{{x}^{3}}\ +\ 5{{x}^{2}}\ -\ 200x\ -\ 500\ =\ 0 \\ & {{x}^{2}}\left( 2x+5 \right)-100\left( 2x+5 \right)=0 \\ & \left( {{x}^{2}}-100 \right)\left( 2x+5 \right)=0 \\ & x=\pm 10,\frac{-5}{2} \end{align}$ Thus, the roots of the equation $2{{x}^{3}}\ +\ 5{{x}^{2}}\ -\ 200x\ -\ 500\ =\ 0$ are $x=\pm 10,\frac{-5}{2}$.
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.