The synthetic division table is:
$$ \begin{array}{c|rrrr}4&15&-52&12&12\\& & 60& 32& \color{black}{176} \\ \hline &\color{blue}{15}&\color{blue}{8}&\color{blue}{44}&\color{orangered}{188} \end{array} $$The solution is:
$$ \frac{ 15x^{3}-52x^{2}+12x+12 }{ x-4 } = \color{blue}{15x^{2}+8x+44} ~+~ \frac{ \color{red}{ 188 } }{ x-4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&15&-52&12&12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 15 }&-52&12&12\\& & & & \\ \hline &\color{orangered}{15}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 15 } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&15&-52&12&12\\& & \color{blue}{60} & & \\ \hline &\color{blue}{15}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -52 } + \color{orangered}{ 60 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}4&15&\color{orangered}{ -52 }&12&12\\& & \color{orangered}{60} & & \\ \hline &15&\color{orangered}{8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 8 } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&15&-52&12&12\\& & 60& \color{blue}{32} & \\ \hline &15&\color{blue}{8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 32 } = \color{orangered}{ 44 } $
$$ \begin{array}{c|rrrr}4&15&-52&\color{orangered}{ 12 }&12\\& & 60& \color{orangered}{32} & \\ \hline &15&8&\color{orangered}{44}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 44 } = \color{blue}{ 176 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&15&-52&12&12\\& & 60& 32& \color{blue}{176} \\ \hline &15&8&\color{blue}{44}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 176 } = \color{orangered}{ 188 } $
$$ \begin{array}{c|rrrr}4&15&-52&12&\color{orangered}{ 12 }\\& & 60& 32& \color{orangered}{176} \\ \hline &\color{blue}{15}&\color{blue}{8}&\color{blue}{44}&\color{orangered}{188} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 15x^{2}+8x+44 } $ with a remainder of $ \color{red}{ 188 } $.