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