Divide using synthetic division $$ ( x^{4}-6x^{3}-40x+33 ) \div ( x-7 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}7&1&-6&0&-40&33\\& & 7& 7& 49& \color{black}{63} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{7}&\color{blue}{9}&\color{orangered}{96} \end{array} $$The solution is:
$$ \dfrac{ x^{4}-6x^{3}-40x+33 }{ x-7 } = \color{blue}{x^{3}+x^{2}+7x+9} ~+~ \dfrac{ \color{red}{ 96 } }{ x-7 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -7 = 0 $ ( $ x = \color{blue}{ 7 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-6&0&-40&33\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}7&\color{orangered}{ 1 }&-6&0&-40&33\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 1 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-6&0&-40&33\\& & \color{blue}{7} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 7 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}7&1&\color{orangered}{ -6 }&0&-40&33\\& & \color{orangered}{7} & & & \\ \hline &1&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 1 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-6&0&-40&33\\& & 7& \color{blue}{7} & & \\ \hline &1&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 7 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}7&1&-6&\color{orangered}{ 0 }&-40&33\\& & 7& \color{orangered}{7} & & \\ \hline &1&1&\color{orangered}{7}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 7 } = \color{blue}{ 49 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-6&0&-40&33\\& & 7& 7& \color{blue}{49} & \\ \hline &1&1&\color{blue}{7}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ 49 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}7&1&-6&0&\color{orangered}{ -40 }&33\\& & 7& 7& \color{orangered}{49} & \\ \hline &1&1&7&\color{orangered}{9}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 9 } = \color{blue}{ 63 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-6&0&-40&33\\& & 7& 7& 49& \color{blue}{63} \\ \hline &1&1&7&\color{blue}{9}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 33 } + \color{orangered}{ 63 } = \color{orangered}{ 96 } $
$$ \begin{array}{c|rrrrr}7&1&-6&0&-40&\color{orangered}{ 33 }\\& & 7& 7& 49& \color{orangered}{63} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{7}&\color{blue}{9}&\color{orangered}{96} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+x^{2}+7x+9 } $ with a remainder of $ \color{red}{ 96 } $.