Divide using synthetic division $$ ( x^{4}-4x^{3}-x^{2}+16x-12 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&1&-4&-1&16&-12\\& & -2& 12& -22& \color{black}{12} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{11}&\color{blue}{-6}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}-4x^{3}-x^{2}+16x-12 }{ x+2 } = \color{blue}{x^{3}-6x^{2}+11x-6} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-4&-1&16&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 1 }&-4&-1&16&-12\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-4&-1&16&-12\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-2&1&\color{orangered}{ -4 }&-1&16&-12\\& & \color{orangered}{-2} & & & \\ \hline &1&\color{orangered}{-6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-4&-1&16&-12\\& & -2& \color{blue}{12} & & \\ \hline &1&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 12 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrr}-2&1&-4&\color{orangered}{ -1 }&16&-12\\& & -2& \color{orangered}{12} & & \\ \hline &1&-6&\color{orangered}{11}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 11 } = \color{blue}{ -22 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-4&-1&16&-12\\& & -2& 12& \color{blue}{-22} & \\ \hline &1&-6&\color{blue}{11}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -22 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-2&1&-4&-1&\color{orangered}{ 16 }&-12\\& & -2& 12& \color{orangered}{-22} & \\ \hline &1&-6&11&\color{orangered}{-6}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-4&-1&16&-12\\& & -2& 12& -22& \color{blue}{12} \\ \hline &1&-6&11&\color{blue}{-6}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 12 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&1&-4&-1&16&\color{orangered}{ -12 }\\& & -2& 12& -22& \color{orangered}{12} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{11}&\color{blue}{-6}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-6x^{2}+11x-6 } $ with a remainder of $ \color{red}{ 0 } $.