Answer
The roots of the polynomial are\[\left\{ 4,\frac{2}{3},-\frac{1}{2}+\frac{\sqrt{3}i}{2},-\frac{1}{2}-\frac{\sqrt{3}i}{2} \right\}\]
Work Step by Step
Consider the given polynomial $f\left( x \right)=3{{x}^{4}}-11{{x}^{3}}-3{{x}^{2}}-6x+8$.
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.
$p=\pm 1,\pm 2,\pm 4,\pm 8$
$q=\pm 1,\pm 2,\pm 3$
Calculate $\frac{p}{q}$.
$\frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{2},\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{8}{3}$
According to the Descartes’s rule of signs, the function $f\left( x \right)=3{{x}^{4}}-11{{x}^{3}}-3{{x}^{2}}-6x+8$ 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 $f\left( -x \right):$
$\begin{align}
& f\left( x \right)=3{{x}^{4}}-11{{x}^{3}}-3{{x}^{2}}-6x+8 \\
& f\left( -x \right)=3{{\left( -x \right)}^{4}}-11{{\left( -x \right)}^{3}}-3{{\left( -x \right)}^{2}}-6\left( -x \right)+8 \\
& =3{{x}^{4}}+11{{x}^{3}}-3{{x}^{2}}+6x+8
\end{align}$
There are two variations in sign. Thus, there are 2 negative real zeros or $2-2=0$ negative real zeros.
Test $x=4$ as a root of the polynomial:
$\begin{align}
& f\left( x \right)=3{{x}^{4}}-11{{x}^{3}}-3{{x}^{2}}-6x+8 \\
& f\left( 4 \right)=3{{\left( 4 \right)}^{4}}-11{{\left( 4 \right)}^{3}}-3{{\left( 4 \right)}^{2}}-6\left( 4 \right)+8 \\
& =0
\end{align}$
Divide the equation $f\left( x \right)$ by $\left( x-4 \right)$.
$\frac{3{{x}^{4}}-11{{x}^{3}}-3{{x}^{2}}-6x+8}{\left( x-4 \right)}=\left( 3x-2 \right)\left( {{x}^{2}}+x+1 \right)$
Thus, the polynomial can be expressed as $f\left( x \right)=\left( x-4 \right)\left( 3x-2 \right)\left( {{x}^{2}}+x+1 \right)$.
Equate $f\left( x \right)=\left( x-4 \right)\left( 3x-2 \right)\left( {{x}^{2}}+x+1 \right)$ to zero.
$\begin{align}
& \left( x-4 \right)\left( 3x-2 \right)\left( {{x}^{2}}+x+1 \right)=0 \\
& \left( x-4 \right)\left( 3x-2 \right)\left( {{\left( x+\frac{1}{2} \right)}^{2}}+\frac{3}{4} \right)=0 \\
& x=4,\frac{2}{3},-\frac{1}{2}+\frac{\sqrt{3}i}{2},-\frac{1}{2}-\frac{\sqrt{3}i}{2}
\end{align}$
The solution of the polynomial is $\left\{ 4,\frac{2}{3},-\frac{1}{2}+\frac{\sqrt{3}i}{2},-\frac{1}{2}-\frac{\sqrt{3}i}{2} \right\}$.
