Divide using synthetic division $$ ( x^{4}+6x^{3}+x^{2}-3x-16 ) \div ( x-6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}6&1&6&1&-3&-16\\& & 6& 72& 438& \color{black}{2610} \\ \hline &\color{blue}{1}&\color{blue}{12}&\color{blue}{73}&\color{blue}{435}&\color{orangered}{2594} \end{array} $$The solution is:
$$ \frac{ x^{4}+6x^{3}+x^{2}-3x-16 }{ x-6 } = \color{blue}{x^{3}+12x^{2}+73x+435} ~+~ \frac{ \color{red}{ 2594 } }{ x-6 } $$Explanation
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}{ 12 } $
$$ \begin{array}{c|rrrrr}6&1&\color{orangered}{ 6 }&1&-3&-16\\& & \color{orangered}{6} & & & \\ \hline &1&\color{orangered}{12}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 12 } = \color{blue}{ 72 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&6&1&-3&-16\\& & 6& \color{blue}{72} & & \\ \hline &1&\color{blue}{12}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 72 } = \color{orangered}{ 73 } $
$$ \begin{array}{c|rrrrr}6&1&6&\color{orangered}{ 1 }&-3&-16\\& & 6& \color{orangered}{72} & & \\ \hline &1&12&\color{orangered}{73}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 73 } = \color{blue}{ 438 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&6&1&-3&-16\\& & 6& 72& \color{blue}{438} & \\ \hline &1&12&\color{blue}{73}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 438 } = \color{orangered}{ 435 } $
$$ \begin{array}{c|rrrrr}6&1&6&1&\color{orangered}{ -3 }&-16\\& & 6& 72& \color{orangered}{438} & \\ \hline &1&12&73&\color{orangered}{435}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 435 } = \color{blue}{ 2610 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&6&1&-3&-16\\& & 6& 72& 438& \color{blue}{2610} \\ \hline &1&12&73&\color{blue}{435}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 2610 } = \color{orangered}{ 2594 } $
$$ \begin{array}{c|rrrrr}6&1&6&1&-3&\color{orangered}{ -16 }\\& & 6& 72& 438& \color{orangered}{2610} \\ \hline &\color{blue}{1}&\color{blue}{12}&\color{blue}{73}&\color{blue}{435}&\color{orangered}{2594} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+12x^{2}+73x+435 } $ with a remainder of $ \color{red}{ 2594 } $.