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