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