Divide using synthetic division $$ ( x^{4}-3x^{3}-3x^{2}-7x+6 ) \div ( x-6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}6&1&-3&-3&-7&6\\& & 6& 18& 90& \color{black}{498} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{15}&\color{blue}{83}&\color{orangered}{504} \end{array} $$The solution is:
$$ \frac{ x^{4}-3x^{3}-3x^{2}-7x+6 }{ x-6 } = \color{blue}{x^{3}+3x^{2}+15x+83} ~+~ \frac{ \color{red}{ 504 } }{ x-6 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-3&-3&-7&6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}6&\color{orangered}{ 1 }&-3&-3&-7&6\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-3&-3&-7&6\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 6 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}6&1&\color{orangered}{ -3 }&-3&-7&6\\& & \color{orangered}{6} & & & \\ \hline &1&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 3 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-3&-3&-7&6\\& & 6& \color{blue}{18} & & \\ \hline &1&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 18 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}6&1&-3&\color{orangered}{ -3 }&-7&6\\& & 6& \color{orangered}{18} & & \\ \hline &1&3&\color{orangered}{15}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 15 } = \color{blue}{ 90 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-3&-3&-7&6\\& & 6& 18& \color{blue}{90} & \\ \hline &1&3&\color{blue}{15}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 90 } = \color{orangered}{ 83 } $
$$ \begin{array}{c|rrrrr}6&1&-3&-3&\color{orangered}{ -7 }&6\\& & 6& 18& \color{orangered}{90} & \\ \hline &1&3&15&\color{orangered}{83}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 83 } = \color{blue}{ 498 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&1&-3&-3&-7&6\\& & 6& 18& 90& \color{blue}{498} \\ \hline &1&3&15&\color{blue}{83}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 498 } = \color{orangered}{ 504 } $
$$ \begin{array}{c|rrrrr}6&1&-3&-3&-7&\color{orangered}{ 6 }\\& & 6& 18& 90& \color{orangered}{498} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{15}&\color{blue}{83}&\color{orangered}{504} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+3x^{2}+15x+83 } $ with a remainder of $ \color{red}{ 504 } $.