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