Divide using synthetic division $$ ( x^{3}-10x^{2}+12x-72 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&-10&12&-72\\& & 3& -21& \color{black}{-27} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{-9}&\color{orangered}{-99} \end{array} $$The solution is:
$$ \frac{ x^{3}-10x^{2}+12x-72 }{ x-3 } = \color{blue}{x^{2}-7x-9} \color{red}{~-~} \frac{ \color{red}{ 99 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-10&12&-72\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&-10&12&-72\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-10&12&-72\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 3 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ -10 }&12&-72\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-10&12&-72\\& & 3& \color{blue}{-21} & \\ \hline &1&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}3&1&-10&\color{orangered}{ 12 }&-72\\& & 3& \color{orangered}{-21} & \\ \hline &1&-7&\color{orangered}{-9}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ -27 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-10&12&-72\\& & 3& -21& \color{blue}{-27} \\ \hline &1&-7&\color{blue}{-9}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -72 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ -99 } $
$$ \begin{array}{c|rrrr}3&1&-10&12&\color{orangered}{ -72 }\\& & 3& -21& \color{orangered}{-27} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{-9}&\color{orangered}{-99} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-7x-9 } $ with a remainder of $ \color{red}{ -99 } $.