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