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