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