Divide using synthetic division $$ ( x^{4}-3x^{3}-3x^{2}-7x+6 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&1&-3&-3&-7&6\\& & 4& 4& 4& \color{black}{-12} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{1}&\color{blue}{-3}&\color{orangered}{-6} \end{array} $$The solution is:
$$ \frac{ x^{4}-3x^{3}-3x^{2}-7x+6 }{ x-4 } = \color{blue}{x^{3}+x^{2}+x-3} \color{red}{~-~} \frac{ \color{red}{ 6 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-3&-3&-7&6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 1 }&-3&-3&-7&6\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-3&-3&-7&6\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 4 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}4&1&\color{orangered}{ -3 }&-3&-7&6\\& & \color{orangered}{4} & & & \\ \hline &1&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-3&-3&-7&6\\& & 4& \color{blue}{4} & & \\ \hline &1&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 4 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}4&1&-3&\color{orangered}{ -3 }&-7&6\\& & 4& \color{orangered}{4} & & \\ \hline &1&1&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-3&-3&-7&6\\& & 4& 4& \color{blue}{4} & \\ \hline &1&1&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 4 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}4&1&-3&-3&\color{orangered}{ -7 }&6\\& & 4& 4& \color{orangered}{4} & \\ \hline &1&1&1&\color{orangered}{-3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&-3&-3&-7&6\\& & 4& 4& 4& \color{blue}{-12} \\ \hline &1&1&1&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}4&1&-3&-3&-7&\color{orangered}{ 6 }\\& & 4& 4& 4& \color{orangered}{-12} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{1}&\color{blue}{-3}&\color{orangered}{-6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+x^{2}+x-3 } $ with a remainder of $ \color{red}{ -6 } $.