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