Divide using synthetic division $$ ( -x^{4}+20x^{2}-19x+12 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&-1&0&20&-19&12\\& & -4& -16& 16& \color{black}{-12} \\ \hline &\color{blue}{-1}&\color{blue}{-4}&\color{blue}{4}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ -x^{4}+20x^{2}-19x+12 }{ x-4 } = \color{blue}{-x^{3}-4x^{2}+4x-3} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&-1&0&20&-19&12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ -1 }&0&20&-19&12\\& & & & & \\ \hline &\color{orangered}{-1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&-1&0&20&-19&12\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{-1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}4&-1&\color{orangered}{ 0 }&20&-19&12\\& & \color{orangered}{-4} & & & \\ \hline &-1&\color{orangered}{-4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&-1&0&20&-19&12\\& & -4& \color{blue}{-16} & & \\ \hline &-1&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}4&-1&0&\color{orangered}{ 20 }&-19&12\\& & -4& \color{orangered}{-16} & & \\ \hline &-1&-4&\color{orangered}{4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 4 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&-1&0&20&-19&12\\& & -4& -16& \color{blue}{16} & \\ \hline &-1&-4&\color{blue}{4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 16 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}4&-1&0&20&\color{orangered}{ -19 }&12\\& & -4& -16& \color{orangered}{16} & \\ \hline &-1&-4&4&\color{orangered}{-3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&-1&0&20&-19&12\\& & -4& -16& 16& \color{blue}{-12} \\ \hline &-1&-4&4&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}4&-1&0&20&-19&\color{orangered}{ 12 }\\& & -4& -16& 16& \color{orangered}{-12} \\ \hline &\color{blue}{-1}&\color{blue}{-4}&\color{blue}{4}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -x^{3}-4x^{2}+4x-3 } $ with a remainder of $ \color{red}{ 0 } $.