Divide using synthetic division $$ ( x^{4}-3x^{3}-3x^{2}-7x+6 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&-3&-3&-7&6\\& & 3& 0& -9& \color{black}{-48} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-3}&\color{blue}{-16}&\color{orangered}{-42} \end{array} $$The solution is:
$$ \frac{ x^{4}-3x^{3}-3x^{2}-7x+6 }{ x-3 } = \color{blue}{x^{3}-3x-16} \color{red}{~-~} \frac{ \color{red}{ 42 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-3&-3&-7&6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&-3&-3&-7&6\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-3&-3&-7&6\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 3 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ -3 }&-3&-7&6\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-3&-3&-7&6\\& & 3& \color{blue}{0} & & \\ \hline &1&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 0 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}3&1&-3&\color{orangered}{ -3 }&-7&6\\& & 3& \color{orangered}{0} & & \\ \hline &1&0&\color{orangered}{-3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-3&-3&-7&6\\& & 3& 0& \color{blue}{-9} & \\ \hline &1&0&\color{blue}{-3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrr}3&1&-3&-3&\color{orangered}{ -7 }&6\\& & 3& 0& \color{orangered}{-9} & \\ \hline &1&0&-3&\color{orangered}{-16}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ -48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-3&-3&-7&6\\& & 3& 0& -9& \color{blue}{-48} \\ \hline &1&0&-3&\color{blue}{-16}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ -42 } $
$$ \begin{array}{c|rrrrr}3&1&-3&-3&-7&\color{orangered}{ 6 }\\& & 3& 0& -9& \color{orangered}{-48} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-3}&\color{blue}{-16}&\color{orangered}{-42} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-3x-16 } $ with a remainder of $ \color{red}{ -42 } $.