Divide using synthetic division $$ ( x^{6}-3x^{5}+6x^{3}-3x^{2}-3x+2 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrrr}-2&1&-3&0&6&-3&-3&2\\& & -2& 10& -20& 28& -50& \color{black}{106} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{10}&\color{blue}{-14}&\color{blue}{25}&\color{blue}{-53}&\color{orangered}{108} \end{array} $$The solution is:
$$ \frac{ x^{6}-3x^{5}+6x^{3}-3x^{2}-3x+2 }{ x+2 } = \color{blue}{x^{5}-5x^{4}+10x^{3}-14x^{2}+25x-53} ~+~ \frac{ \color{red}{ 108 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-2}&1&-3&0&6&-3&-3&2\\& & & & & & & \\ \hline &&&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrrr}-2&\color{orangered}{ 1 }&-3&0&6&-3&-3&2\\& & & & & & & \\ \hline &\color{orangered}{1}&&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-2}&1&-3&0&6&-3&-3&2\\& & \color{blue}{-2} & & & & & \\ \hline &\color{blue}{1}&&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrrrr}-2&1&\color{orangered}{ -3 }&0&6&-3&-3&2\\& & \color{orangered}{-2} & & & & & \\ \hline &1&\color{orangered}{-5}&&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-2}&1&-3&0&6&-3&-3&2\\& & -2& \color{blue}{10} & & & & \\ \hline &1&\color{blue}{-5}&&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 10 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrrr}-2&1&-3&\color{orangered}{ 0 }&6&-3&-3&2\\& & -2& \color{orangered}{10} & & & & \\ \hline &1&-5&\color{orangered}{10}&&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 10 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-2}&1&-3&0&6&-3&-3&2\\& & -2& 10& \color{blue}{-20} & & & \\ \hline &1&-5&\color{blue}{10}&&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrrrrr}-2&1&-3&0&\color{orangered}{ 6 }&-3&-3&2\\& & -2& 10& \color{orangered}{-20} & & & \\ \hline &1&-5&10&\color{orangered}{-14}&&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-2}&1&-3&0&6&-3&-3&2\\& & -2& 10& -20& \color{blue}{28} & & \\ \hline &1&-5&10&\color{blue}{-14}&&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 28 } = \color{orangered}{ 25 } $
$$ \begin{array}{c|rrrrrrr}-2&1&-3&0&6&\color{orangered}{ -3 }&-3&2\\& & -2& 10& -20& \color{orangered}{28} & & \\ \hline &1&-5&10&-14&\color{orangered}{25}&& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 25 } = \color{blue}{ -50 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-2}&1&-3&0&6&-3&-3&2\\& & -2& 10& -20& 28& \color{blue}{-50} & \\ \hline &1&-5&10&-14&\color{blue}{25}&& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -50 \right) } = \color{orangered}{ -53 } $
$$ \begin{array}{c|rrrrrrr}-2&1&-3&0&6&-3&\color{orangered}{ -3 }&2\\& & -2& 10& -20& 28& \color{orangered}{-50} & \\ \hline &1&-5&10&-14&25&\color{orangered}{-53}& \end{array} $$Step 12 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -53 \right) } = \color{blue}{ 106 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-2}&1&-3&0&6&-3&-3&2\\& & -2& 10& -20& 28& -50& \color{blue}{106} \\ \hline &1&-5&10&-14&25&\color{blue}{-53}& \end{array} $$Step 13 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 106 } = \color{orangered}{ 108 } $
$$ \begin{array}{c|rrrrrrr}-2&1&-3&0&6&-3&-3&\color{orangered}{ 2 }\\& & -2& 10& -20& 28& -50& \color{orangered}{106} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{10}&\color{blue}{-14}&\color{blue}{25}&\color{blue}{-53}&\color{orangered}{108} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{5}-5x^{4}+10x^{3}-14x^{2}+25x-53 } $ with a remainder of $ \color{red}{ 108 } $.