The synthetic division table is:
$$ \begin{array}{c|rrr}4&3&0&-3\\& & 12& \color{black}{48} \\ \hline &\color{blue}{3}&\color{blue}{12}&\color{orangered}{45} \end{array} $$The solution is:
$$ \frac{ 3x^{2}-3 }{ x-4 } = \color{blue}{3x+12} ~+~ \frac{ \color{red}{ 45 } }{ x-4 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrr}\color{blue}{4}&3&0&-3\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}4&\color{orangered}{ 3 }&0&-3\\& & & \\ \hline &\color{orangered}{3}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 3 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrr}\color{blue}{4}&3&0&-3\\& & \color{blue}{12} & \\ \hline &\color{blue}{3}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrr}4&3&\color{orangered}{ 0 }&-3\\& & \color{orangered}{12} & \\ \hline &3&\color{orangered}{12}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 12 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrr}\color{blue}{4}&3&0&-3\\& & 12& \color{blue}{48} \\ \hline &3&\color{blue}{12}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 48 } = \color{orangered}{ 45 } $
$$ \begin{array}{c|rrr}4&3&0&\color{orangered}{ -3 }\\& & 12& \color{orangered}{48} \\ \hline &\color{blue}{3}&\color{blue}{12}&\color{orangered}{45} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x+12 } $ with a remainder of $ \color{red}{ 45 } $.