Answer
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 $x=1-\sqrt{3}$.
