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