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: 26


The solutions of the equation are $x=-3,x=\frac{1}{2},x=1+\sqrt{3}$ ,and $x=1-\sqrt{3}$.

Work Step by Step

To solve the polynomial equation $2{{x}^{4}}+{{x}^{3}}-17{{x}^{2}}-4x+6=0$. By the rational root theorem, ${{a}_{0}}=6\text{ , }{{\text{a}}_{n}}=2$ The factors of ${{a}_{0}}$ are: 1,2,3,6 and the factors of ${{a}_{n}}$ are: 1,2 Possible rational roots are: $\pm \frac{1,2,3,6}{1,2}$. $\frac{-3}{1}$ is a root of the expression, so factor out $x+3$. Compute $\frac{2{{x}^{4}}+{{x}^{3}}-17{{x}^{2}}-4x+6}{x+3}$ to get the rest of the equation: $2{{x}^{3}}-5{{x}^{2}}-2x+2$. $\left( x+3 \right)\left( 2{{x}^{3}}-5{{x}^{2}}-2x+2 \right)=0$ $x=-3$ , $x=\frac{1}{2}$ , $x=1+\sqrt{3}$ $x=1-\sqrt{3}$. The solutions of the equations are $x=-3,x=\frac{1}{2},x=1+\sqrt{3}$ ,and $x=1-\sqrt{3}$.
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