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