The synthetic division table is:
$$ \begin{array}{c|rrrr}1&3&8&3&-2\\& & 3& 11& \color{black}{14} \\ \hline &\color{blue}{3}&\color{blue}{11}&\color{blue}{14}&\color{orangered}{12} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+8x^{2}+3x-2 }{ x-1 } = \color{blue}{3x^{2}+11x+14} ~+~ \frac{ \color{red}{ 12 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&8&3&-2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 3 }&8&3&-2\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&8&3&-2\\& & \color{blue}{3} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 3 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrr}1&3&\color{orangered}{ 8 }&3&-2\\& & \color{orangered}{3} & & \\ \hline &3&\color{orangered}{11}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 11 } = \color{blue}{ 11 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&8&3&-2\\& & 3& \color{blue}{11} & \\ \hline &3&\color{blue}{11}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 11 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrr}1&3&8&\color{orangered}{ 3 }&-2\\& & 3& \color{orangered}{11} & \\ \hline &3&11&\color{orangered}{14}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 14 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&8&3&-2\\& & 3& 11& \color{blue}{14} \\ \hline &3&11&\color{blue}{14}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 14 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}1&3&8&3&\color{orangered}{ -2 }\\& & 3& 11& \color{orangered}{14} \\ \hline &\color{blue}{3}&\color{blue}{11}&\color{blue}{14}&\color{orangered}{12} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+11x+14 } $ with a remainder of $ \color{red}{ 12 } $.