Divide using synthetic division $$ ( x^{3}-9x^{2}+20x-12 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&1&-9&20&-12\\& & 4& -20& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{0}&\color{orangered}{-12} \end{array} $$The solution is:
$$ \frac{ x^{3}-9x^{2}+20x-12 }{ x-4 } = \color{blue}{x^{2}-5x} \color{red}{~-~} \frac{ \color{red}{ 12 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-9&20&-12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 1 }&-9&20&-12\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-9&20&-12\\& & \color{blue}{4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 4 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}4&1&\color{orangered}{ -9 }&20&-12\\& & \color{orangered}{4} & & \\ \hline &1&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-9&20&-12\\& & 4& \color{blue}{-20} & \\ \hline &1&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}4&1&-9&\color{orangered}{ 20 }&-12\\& & 4& \color{orangered}{-20} & \\ \hline &1&-5&\color{orangered}{0}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-9&20&-12\\& & 4& -20& \color{blue}{0} \\ \hline &1&-5&\color{blue}{0}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 0 } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}4&1&-9&20&\color{orangered}{ -12 }\\& & 4& -20& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{0}&\color{orangered}{-12} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-5x } $ with a remainder of $ \color{red}{ -12 } $.