Answer
The solutions of the equation are $x=1$ , $x=\frac{1}{3}$ , and $x=\frac{1}{2}$.
Work Step by Step
Consider the polynomial equation $f\left( x \right)=6{{x}^{3}}-11{{x}^{2}}+6x-1$.
Use the rational root theorem to observe:
${{a}_{0}}=1\text{ , }{{\text{a}}_{n}}=6$
Factors of ${{a}_{0}}$ are: 1 and the factors of ${{a}_{n}}$ are: 1, 2, 3, 6.
The possible roots are given as: $\pm \frac{1}{1,2,3,6}$.
$\frac{1}{1}$ is a root of the expression, so factor out is $x-1$.
Compute $\frac{6{{x}^{3}}\ -\ 11{{x}^{2}}\ +\ 6x\ -\ 1}{x\ -\ 1}$ to get the rest of the equation: $6{{x}^{2}}\ -\ 5x\ +\ 1$.
$\begin{align}
& \left( x\ -\ 1 \right)\ \left( 6{{x}^{2}}\ -\ 5x\ +\ 1 \right)=0 \\
& \left( x-1 \right)\left( 2x-1 \right)\left( 3x-1 \right)=0 \\
& x=1,\frac{1}{2},\frac{1}{3}
\end{align}$
Hence, the roots of the polynomial equation $f\left( x \right)=6{{x}^{3}}-11{{x}^{2}}+6x-1$ are $x=1,\frac{1}{2},\frac{1}{3}$.
