The synthetic division table is:
$$ \begin{array}{c|rrrr}1&3&-4&-17&6\\& & 3& -1& \color{black}{-18} \\ \hline &\color{blue}{3}&\color{blue}{-1}&\color{blue}{-18}&\color{orangered}{-12} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-4x^{2}-17x+6 }{ x-1 } = \color{blue}{3x^{2}-x-18} \color{red}{~-~} \frac{ \color{red}{ 12 } }{ x-1 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&-4&-17&6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 3 }&-4&-17&6\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&-4&-17&6\\& & \color{blue}{3} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 3 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}1&3&\color{orangered}{ -4 }&-17&6\\& & \color{orangered}{3} & & \\ \hline &3&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -1 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&-4&-17&6\\& & 3& \color{blue}{-1} & \\ \hline &3&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ -18 } $
$$ \begin{array}{c|rrrr}1&3&-4&\color{orangered}{ -17 }&6\\& & 3& \color{orangered}{-1} & \\ \hline &3&-1&\color{orangered}{-18}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -18 \right) } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&-4&-17&6\\& & 3& -1& \color{blue}{-18} \\ \hline &3&-1&\color{blue}{-18}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}1&3&-4&-17&\color{orangered}{ 6 }\\& & 3& -1& \color{orangered}{-18} \\ \hline &\color{blue}{3}&\color{blue}{-1}&\color{blue}{-18}&\color{orangered}{-12} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-x-18 } $ with a remainder of $ \color{red}{ -12 } $.