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