Divide using synthetic division $$ ( x^{4}-6x^{3}+x^{2}-3x-16 ) \div ( x-6 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}6&1&-6&1&-3&-16\\& & 6& 0& 6& \color{black}{18} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{1}&\color{blue}{3}&\color{orangered}{2} \end{array} $$The solution is:
$$ \dfrac{ x^{4}-6x^{3}+x^{2}-3x-16 }{ x-6 } = \color{blue}{x^{3}+x+3} ~+~ \dfrac{ \color{red}{ 2 } }{ x-6 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-6&1&-3&-16\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}6&\color{orangered}{ 1 }&-6&1&-3&-16\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-6&1&-3&-16\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 6 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}6&1&\color{orangered}{ -6 }&1&-3&-16\\& & \color{orangered}{6} & & & \\ \hline &1&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-6&1&-3&-16\\& & 6& \color{blue}{0} & & \\ \hline &1&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 0 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}6&1&-6&\color{orangered}{ 1 }&-3&-16\\& & 6& \color{orangered}{0} & & \\ \hline &1&0&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 1 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-6&1&-3&-16\\& & 6& 0& \color{blue}{6} & \\ \hline &1&0&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 6 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}6&1&-6&1&\color{orangered}{ -3 }&-16\\& & 6& 0& \color{orangered}{6} & \\ \hline &1&0&1&\color{orangered}{3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 3 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-6&1&-3&-16\\& & 6& 0& 6& \color{blue}{18} \\ \hline &1&0&1&\color{blue}{3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 18 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}6&1&-6&1&-3&\color{orangered}{ -16 }\\& & 6& 0& 6& \color{orangered}{18} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{1}&\color{blue}{3}&\color{orangered}{2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+x+3 } $ with a remainder of $ \color{red}{ 2 } $.