The synthetic division table is:
$$ \begin{array}{c|rr}3&8&0\\& & \color{black}{24} \\ \hline &\color{blue}{8}&\color{orangered}{24} \end{array} $$The solution is:
$$ \frac{ 8x }{ x-3 } = \color{blue}{8} ~+~ \frac{ \color{red}{ 24 } }{ x-3 } $$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.
$$ \begin{array}{c|rr}\color{blue}{3}&8&0\\& & \\ \hline && \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rr}3&\color{orangered}{ 8 }&0\\& & \\ \hline &\color{orangered}{8}& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 8 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rr}\color{blue}{3}&8&0\\& & \color{blue}{24} \\ \hline &\color{blue}{8}& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 24 } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rr}3&8&\color{orangered}{ 0 }\\& & \color{orangered}{24} \\ \hline &\color{blue}{8}&\color{orangered}{24} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 8 } $ with a remainder of $ \color{red}{ 24 } $.