Divide using synthetic division $$ ( x^{5}+x^{4}-8x^{3}-7x^{2}+7x+33 ) \div ( x+3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-3&1&1&-8&-7&7&33\\& & -3& 6& 6& 3& \color{black}{-30} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{-2}&\color{blue}{-1}&\color{blue}{10}&\color{orangered}{3} \end{array} $$The solution is:
$$ \dfrac{ x^{5}+x^{4}-8x^{3}-7x^{2}+7x+33 }{ x+3 } = \color{blue}{x^{4}-2x^{3}-2x^{2}-x+10} ~+~ \dfrac{ \color{red}{ 3 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&1&-8&-7&7&33\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-3&\color{orangered}{ 1 }&1&-8&-7&7&33\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&1&-8&-7&7&33\\& & \color{blue}{-3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}-3&1&\color{orangered}{ 1 }&-8&-7&7&33\\& & \color{orangered}{-3} & & & & \\ \hline &1&\color{orangered}{-2}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&1&-8&-7&7&33\\& & -3& \color{blue}{6} & & & \\ \hline &1&\color{blue}{-2}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 6 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}-3&1&1&\color{orangered}{ -8 }&-7&7&33\\& & -3& \color{orangered}{6} & & & \\ \hline &1&-2&\color{orangered}{-2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&1&-8&-7&7&33\\& & -3& 6& \color{blue}{6} & & \\ \hline &1&-2&\color{blue}{-2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 6 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}-3&1&1&-8&\color{orangered}{ -7 }&7&33\\& & -3& 6& \color{orangered}{6} & & \\ \hline &1&-2&-2&\color{orangered}{-1}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&1&-8&-7&7&33\\& & -3& 6& 6& \color{blue}{3} & \\ \hline &1&-2&-2&\color{blue}{-1}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 3 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrr}-3&1&1&-8&-7&\color{orangered}{ 7 }&33\\& & -3& 6& 6& \color{orangered}{3} & \\ \hline &1&-2&-2&-1&\color{orangered}{10}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 10 } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&1&-8&-7&7&33\\& & -3& 6& 6& 3& \color{blue}{-30} \\ \hline &1&-2&-2&-1&\color{blue}{10}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 33 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}-3&1&1&-8&-7&7&\color{orangered}{ 33 }\\& & -3& 6& 6& 3& \color{orangered}{-30} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{-2}&\color{blue}{-1}&\color{blue}{10}&\color{orangered}{3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-2x^{3}-2x^{2}-x+10 } $ with a remainder of $ \color{red}{ 3 } $.