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