Answer
The roots of the polynomial equation are\[\left\{ -\frac{1}{2},\frac{1}{2}+\frac{\sqrt{17}}{2},\frac{1}{2}-\frac{\sqrt{17}}{2} \right\}\]
Work Step by Step
Consider the given polynomial $2{{x}^{3}}-{{x}^{2}}-9x-4=0$.
Determine values of p and q where p are the factors of the constant in the polynomial and q are the factors of the leading coefficient of the polynomial.
$p=\pm 1,\pm 2,\pm 4$
$q=\pm 1,\pm 2$
Calculate $\frac{p}{q}$.
$\frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm \frac{1}{2}$
According to the Descartes’ s rule of signs, the function $2{{x}^{3}}-{{x}^{2}}-9x-4=0$ has two sign changes. Since the function $f\left( x \right)$ has two variations in sign, the function $f\left( x \right)$ has two or zero positive real root.
Further, evaluate $f\left( -x \right):$
$\begin{align}
& f\left( x \right)=2{{x}^{3}}-{{x}^{2}}-9x-4 \\
& f\left( -x \right)=2{{\left( -x \right)}^{3}}-{{\left( -x \right)}^{2}}-9\left( -x \right)-4 \\
& =-2{{x}^{3}}-{{x}^{2}}+9x-4
\end{align}$
There are two variations in sign. Thus, there are 2 negative real zeros or $2-2=0$ negative real zeros.
Test $x=\frac{-1}{2}$ as a root of the polynomial:
$\begin{align}
& f\left( x \right)=2{{x}^{3}}-{{x}^{2}}-9x-4 \\
& f\left( -\frac{1}{2} \right)={{\left( -\frac{1}{2} \right)}^{3}}+12{{\left( -\frac{1}{2} \right)}^{2}}+21\left( -\frac{1}{2} \right)+10 \\
& =0
\end{align}$
Divide the equation $f\left( x \right)$ by $\left( x+\frac{1}{2} \right)$.
$\frac{2{{x}^{3}}-{{x}^{2}}-9x-4}{x+\frac{1}{2}}={{x}^{2}}-x-4$
Thus, the polynomial can be expressed as $f\left( x \right)=\left( x+\frac{1}{2} \right)\left( {{x}^{2}}-x-4 \right)$.
Equate $f\left( x \right)=\left( x+\frac{1}{2} \right)\left( {{x}^{2}}-x-4 \right)$ to zero.
$\begin{align}
& \left( x+\frac{1}{2} \right)\left( {{x}^{2}}-x-4 \right)=0 \\
& \left( x+\frac{1}{2} \right)\left( {{\left( x-\frac{1}{2} \right)}^{2}}-\frac{17}{4} \right)=0 \\
& x=-\frac{1}{2},\frac{1}{2}+\frac{\sqrt{17}}{2},\frac{1}{2}-\frac{\sqrt{17}}{2}
\end{align}$
The solution of the polynomial is $\left\{ -\frac{1}{2},\frac{1}{2}+\frac{\sqrt{17}}{2},\frac{1}{2}-\frac{\sqrt{17}}{2} \right\}$.
