Divide using synthetic division $$ ( x^{7}-4x^{5}+3x-33 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrrrr}-2&1&0&-4&0&0&0&3&-33\\& & -2& 4& 0& 0& 0& 0& \color{black}{-6} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{0}&\color{blue}{0}&\color{blue}{0}&\color{blue}{0}&\color{blue}{3}&\color{orangered}{-39} \end{array} $$The solution is:
$$ \frac{ x^{7}-4x^{5}+3x-33 }{ x+2 } = \color{blue}{x^{6}-2x^{5}+3} \color{red}{~-~} \frac{ \color{red}{ 39 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrrrrr}\color{blue}{-2}&1&0&-4&0&0&0&3&-33\\& & & & & & & & \\ \hline &&&&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrrrr}-2&\color{orangered}{ 1 }&0&-4&0&0&0&3&-33\\& & & & & & & & \\ \hline &\color{orangered}{1}&&&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrrrrr}\color{blue}{-2}&1&0&-4&0&0&0&3&-33\\& & \color{blue}{-2} & & & & & & \\ \hline &\color{blue}{1}&&&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrrrr}-2&1&\color{orangered}{ 0 }&-4&0&0&0&3&-33\\& & \color{orangered}{-2} & & & & & & \\ \hline &1&\color{orangered}{-2}&&&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrrrr}\color{blue}{-2}&1&0&-4&0&0&0&3&-33\\& & -2& \color{blue}{4} & & & & & \\ \hline &1&\color{blue}{-2}&&&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 4 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrrr}-2&1&0&\color{orangered}{ -4 }&0&0&0&3&-33\\& & -2& \color{orangered}{4} & & & & & \\ \hline &1&-2&\color{orangered}{0}&&&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrrr}\color{blue}{-2}&1&0&-4&0&0&0&3&-33\\& & -2& 4& \color{blue}{0} & & & & \\ \hline &1&-2&\color{blue}{0}&&&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrrr}-2&1&0&-4&\color{orangered}{ 0 }&0&0&3&-33\\& & -2& 4& \color{orangered}{0} & & & & \\ \hline &1&-2&0&\color{orangered}{0}&&&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrrr}\color{blue}{-2}&1&0&-4&0&0&0&3&-33\\& & -2& 4& 0& \color{blue}{0} & & & \\ \hline &1&-2&0&\color{blue}{0}&&&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrrr}-2&1&0&-4&0&\color{orangered}{ 0 }&0&3&-33\\& & -2& 4& 0& \color{orangered}{0} & & & \\ \hline &1&-2&0&0&\color{orangered}{0}&&& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrrr}\color{blue}{-2}&1&0&-4&0&0&0&3&-33\\& & -2& 4& 0& 0& \color{blue}{0} & & \\ \hline &1&-2&0&0&\color{blue}{0}&&& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrrr}-2&1&0&-4&0&0&\color{orangered}{ 0 }&3&-33\\& & -2& 4& 0& 0& \color{orangered}{0} & & \\ \hline &1&-2&0&0&0&\color{orangered}{0}&& \end{array} $$Step 12 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrrr}\color{blue}{-2}&1&0&-4&0&0&0&3&-33\\& & -2& 4& 0& 0& 0& \color{blue}{0} & \\ \hline &1&-2&0&0&0&\color{blue}{0}&& \end{array} $$Step 13 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 0 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrrrr}-2&1&0&-4&0&0&0&\color{orangered}{ 3 }&-33\\& & -2& 4& 0& 0& 0& \color{orangered}{0} & \\ \hline &1&-2&0&0&0&0&\color{orangered}{3}& \end{array} $$Step 14 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrrrrr}\color{blue}{-2}&1&0&-4&0&0&0&3&-33\\& & -2& 4& 0& 0& 0& 0& \color{blue}{-6} \\ \hline &1&-2&0&0&0&0&\color{blue}{3}& \end{array} $$Step 15 : Add down last column: $ \color{orangered}{ -33 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -39 } $
$$ \begin{array}{c|rrrrrrrr}-2&1&0&-4&0&0&0&3&\color{orangered}{ -33 }\\& & -2& 4& 0& 0& 0& 0& \color{orangered}{-6} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{0}&\color{blue}{0}&\color{blue}{0}&\color{blue}{0}&\color{blue}{3}&\color{orangered}{-39} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{6}-2x^{5}+3 } $ with a remainder of $ \color{red}{ -39 } $.