Answer
The roots of the equation are:
\[\left\{ \frac{2}{3},1-\sqrt{5},1+\sqrt{5} \right\}\]
Consider the given polynomial:
\[3{{x}^{3}}-8{{x}^{2}}-8x+8=0\]
Work Step by Step
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 $3{{x}^{3}}-8{{x}^{2}}-8x+8=0$ 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 $f\left( -x \right):$
$\begin{align}
& f\left( x \right)=3{{x}^{3}}-8{{x}^{2}}-8x+8 \\
& f\left( -x \right)=3{{\left( -x \right)}^{3}}-8{{\left( -x \right)}^{2}}-8\left( -x \right)+8 \\
& =3{{x}^{3}}-8{{x}^{2}}+8x+8
\end{align}$
There are two variations in sign. Thus, there are 2 negative real zeros or $2-2=0$ negative real zeros.
The function has only one sign change, thus the function has one negative real zero.
Test $x=\frac{2}{3}$ as a root of the polynomial:
$\begin{align}
& f\left( x \right)=3{{x}^{3}}-8{{x}^{2}}-8x+8 \\
& f\left( \frac{2}{3} \right)=3{{\left( \frac{2}{3} \right)}^{3}}-8{{\left( \frac{2}{3} \right)}^{2}}-8\left( \frac{2}{3} \right)+8 \\
& =0
\end{align}$
Thus, $\left( x-\frac{2}{3} \right)$ is a factor of the polynomial.
Divide the polynomial by $\left( x-\frac{2}{3} \right)$
$\frac{\left( 3{{x}^{3}}-8{{x}^{2}}-8x+8 \right)}{\left( x-\frac{2}{3} \right)}={{x}^{2}}-2x-4$
Thus, express the function as $f\left( x \right)=\left( x-\frac{2}{3} \right)\left( {{x}^{2}}-2x-4 \right)$.
Equate $f\left( x \right)=\left( x-\frac{2}{3} \right)\left( {{x}^{2}}-2x-4 \right)$ to zero.
$\begin{align}
& \left( x-\frac{2}{3} \right)\left( {{x}^{2}}-2x-4 \right)=0 \\
& \left( x-\frac{2}{3} \right)\left( {{\left( x-1 \right)}^{2}}-5 \right)=0 \\
& x=\frac{2}{3},1-\sqrt{5},1+\sqrt{5}
\end{align}$
The solution of the polynomial is $\left\{ \frac{2}{3},1-\sqrt{5},1+\sqrt{5} \right\}$.
