Divide using synthetic division $$ ( x^{4}+2x^{3}-12x^{2}-11x+6 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&1&2&-12&-11&6\\& & -2& 0& 24& \color{black}{-26} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-12}&\color{blue}{13}&\color{orangered}{-20} \end{array} $$The solution is:
$$ \frac{ x^{4}+2x^{3}-12x^{2}-11x+6 }{ x+2 } = \color{blue}{x^{3}-12x+13} \color{red}{~-~} \frac{ \color{red}{ 20 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&2&-12&-11&6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 1 }&2&-12&-11&6\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&2&-12&-11&6\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&1&\color{orangered}{ 2 }&-12&-11&6\\& & \color{orangered}{-2} & & & \\ \hline &1&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&2&-12&-11&6\\& & -2& \color{blue}{0} & & \\ \hline &1&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 0 } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}-2&1&2&\color{orangered}{ -12 }&-11&6\\& & -2& \color{orangered}{0} & & \\ \hline &1&0&\color{orangered}{-12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&2&-12&-11&6\\& & -2& 0& \color{blue}{24} & \\ \hline &1&0&\color{blue}{-12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 24 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrr}-2&1&2&-12&\color{orangered}{ -11 }&6\\& & -2& 0& \color{orangered}{24} & \\ \hline &1&0&-12&\color{orangered}{13}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 13 } = \color{blue}{ -26 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&2&-12&-11&6\\& & -2& 0& 24& \color{blue}{-26} \\ \hline &1&0&-12&\color{blue}{13}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -26 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrr}-2&1&2&-12&-11&\color{orangered}{ 6 }\\& & -2& 0& 24& \color{orangered}{-26} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-12}&\color{blue}{13}&\color{orangered}{-20} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-12x+13 } $ with a remainder of $ \color{red}{ -20 } $.