Divide using synthetic division $$ ( x^{3}-10x^{2}+12x-72 ) \div ( x-6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}6&1&-10&12&-72\\& & 6& -24& \color{black}{-72} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-12}&\color{orangered}{-144} \end{array} $$The solution is:
$$ \frac{ x^{3}-10x^{2}+12x-72 }{ x-6 } = \color{blue}{x^{2}-4x-12} \color{red}{~-~} \frac{ \color{red}{ 144 } }{ x-6 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -6 = 0 $ ( $ x = \color{blue}{ 6 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{6}&1&-10&12&-72\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}6&\color{orangered}{ 1 }&-10&12&-72\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 1 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&1&-10&12&-72\\& & \color{blue}{6} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 6 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}6&1&\color{orangered}{ -10 }&12&-72\\& & \color{orangered}{6} & & \\ \hline &1&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&1&-10&12&-72\\& & 6& \color{blue}{-24} & \\ \hline &1&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}6&1&-10&\color{orangered}{ 12 }&-72\\& & 6& \color{orangered}{-24} & \\ \hline &1&-4&\color{orangered}{-12}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ -72 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&1&-10&12&-72\\& & 6& -24& \color{blue}{-72} \\ \hline &1&-4&\color{blue}{-12}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -72 } + \color{orangered}{ \left( -72 \right) } = \color{orangered}{ -144 } $
$$ \begin{array}{c|rrrr}6&1&-10&12&\color{orangered}{ -72 }\\& & 6& -24& \color{orangered}{-72} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-12}&\color{orangered}{-144} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-4x-12 } $ with a remainder of $ \color{red}{ -144 } $.