Answer
The solution is, $\underline{x=1,-4\,\,\text{or}\,\,\frac{1}{3}}$.
Work Step by Step
Consider the given equation,
$\begin{align}
& 3{{x}^{3}}+8{{x}^{2}}-15x+4=0 \\
& 3{{x}^{3}}-3{{x}^{2}}+11{{x}^{2}}-11x-4x+4=0 \\
& 3{{x}^{2}}\left( x-1 \right)+11x\left( x-1 \right)-4\left( x-1 \right)=0 \\
& \left( x-1 \right)\left( 3{{x}^{2}}+11x-4 \right)=0
\end{align}$
It can be further simplified as below:
$\begin{align}
& \left( x-1 \right)\left( 3{{x}^{2}}+12x-x-4 \right)=0 \\
& \left( x-1 \right)\left\{ 3x\left( x+4 \right)-\left( x+4 \right) \right\}=0 \\
& \left( x-1 \right)\left( x+4 \right)\left( 3x-1 \right)=0
\end{align}$
Therefore
$x=1,-4\,\,\text{or}\,\,\frac{1}{3}$
The solutions of the equations are $x=1,-4\,\,\text{or}\,\,\frac{1}{3}$.
