The synthetic division table is:
$$ \begin{array}{c|rrr}3&2&-10&8\\& & 6& \color{black}{-12} \\ \hline &\color{blue}{2}&\color{blue}{-4}&\color{orangered}{-4} \end{array} $$The solution is:
$$ \frac{ 2x^{2}-10x+8 }{ x-3 } = \color{blue}{2x-4} \color{red}{~-~} \frac{ \color{red}{ 4 } }{ x-3 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrr}\color{blue}{3}&2&-10&8\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}3&\color{orangered}{ 2 }&-10&8\\& & & \\ \hline &\color{orangered}{2}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&2&-10&8\\& & \color{blue}{6} & \\ \hline &\color{blue}{2}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 6 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrr}3&2&\color{orangered}{ -10 }&8\\& & \color{orangered}{6} & \\ \hline &2&\color{orangered}{-4}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&2&-10&8\\& & 6& \color{blue}{-12} \\ \hline &2&\color{blue}{-4}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrr}3&2&-10&\color{orangered}{ 8 }\\& & 6& \color{orangered}{-12} \\ \hline &\color{blue}{2}&\color{blue}{-4}&\color{orangered}{-4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x-4 } $ with a remainder of $ \color{red}{ -4 } $.