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