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