The synthetic division table is:
$$ \begin{array}{c|rrrr}0&3&-9&2&0\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{3}&\color{blue}{-9}&\color{blue}{2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-9x^{2}+2x }{ x } = \color{blue}{3x^{2}-9x+2} $$Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&3&-9&2&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 3 }&-9&2&0\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 3 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&3&-9&2&0\\& & \color{blue}{0} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 0 } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}0&3&\color{orangered}{ -9 }&2&0\\& & \color{orangered}{0} & & \\ \hline &3&\color{orangered}{-9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&3&-9&2&0\\& & 0& \color{blue}{0} & \\ \hline &3&\color{blue}{-9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 0 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}0&3&-9&\color{orangered}{ 2 }&0\\& & 0& \color{orangered}{0} & \\ \hline &3&-9&\color{orangered}{2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 2 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&3&-9&2&0\\& & 0& 0& \color{blue}{0} \\ \hline &3&-9&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}0&3&-9&2&\color{orangered}{ 0 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{3}&\color{blue}{-9}&\color{blue}{2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-9x+2 } $ with a remainder of $ \color{red}{ 0 } $.