Divide using synthetic division $$ ( x^{3}-8x^{2}+4x+48 ) \div ( x-4 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&1&-8&4&48\\& & 4& -16& \color{black}{-48} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-12}&\color{orangered}{0} \end{array} $$The solution is:
$$ \dfrac{ x^{3}-8x^{2}+4x+48 }{ x-4 } = \color{blue}{x^{2}-4x-12} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-8&4&48\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 1 }&-8&4&48\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-8&4&48\\& & \color{blue}{4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 4 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}4&1&\color{orangered}{ -8 }&4&48\\& & \color{orangered}{4} & & \\ \hline &1&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-8&4&48\\& & 4& \color{blue}{-16} & \\ \hline &1&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}4&1&-8&\color{orangered}{ 4 }&48\\& & 4& \color{orangered}{-16} & \\ \hline &1&-4&\color{orangered}{-12}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ -48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-8&4&48\\& & 4& -16& \color{blue}{-48} \\ \hline &1&-4&\color{blue}{-12}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 48 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}4&1&-8&4&\color{orangered}{ 48 }\\& & 4& -16& \color{orangered}{-48} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-12}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-4x-12 } $ with a remainder of $ \color{red}{ 0 } $.