Divide using synthetic division $$ ( -x^{5}-5x^{3}-x^{2}+2 ) \div ( 3x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}1&-1&0&-5&-1&0&2\\& & -1& -1& -6& -7& \color{black}{-7} \\ \hline &\color{blue}{-1}&\color{blue}{-1}&\color{blue}{-6}&\color{blue}{-7}&\color{blue}{-7}&\color{orangered}{-5} \end{array} $$The solution is:
$$ \frac{ -x^{5}-5x^{3}-x^{2}+2 }{ x-1 } = \color{blue}{-x^{4}-x^{3}-6x^{2}-7x-7} \color{red}{~-~} \frac{ \color{red}{ 5 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&-1&0&-5&-1&0&2\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}1&\color{orangered}{ -1 }&0&-5&-1&0&2\\& & & & & & \\ \hline &\color{orangered}{-1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&-1&0&-5&-1&0&2\\& & \color{blue}{-1} & & & & \\ \hline &\color{blue}{-1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}1&-1&\color{orangered}{ 0 }&-5&-1&0&2\\& & \color{orangered}{-1} & & & & \\ \hline &-1&\color{orangered}{-1}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -1 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&-1&0&-5&-1&0&2\\& & -1& \color{blue}{-1} & & & \\ \hline &-1&\color{blue}{-1}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrrr}1&-1&0&\color{orangered}{ -5 }&-1&0&2\\& & -1& \color{orangered}{-1} & & & \\ \hline &-1&-1&\color{orangered}{-6}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&-1&0&-5&-1&0&2\\& & -1& -1& \color{blue}{-6} & & \\ \hline &-1&-1&\color{blue}{-6}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrrr}1&-1&0&-5&\color{orangered}{ -1 }&0&2\\& & -1& -1& \color{orangered}{-6} & & \\ \hline &-1&-1&-6&\color{orangered}{-7}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -7 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&-1&0&-5&-1&0&2\\& & -1& -1& -6& \color{blue}{-7} & \\ \hline &-1&-1&-6&\color{blue}{-7}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrrr}1&-1&0&-5&-1&\color{orangered}{ 0 }&2\\& & -1& -1& -6& \color{orangered}{-7} & \\ \hline &-1&-1&-6&-7&\color{orangered}{-7}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -7 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&-1&0&-5&-1&0&2\\& & -1& -1& -6& -7& \color{blue}{-7} \\ \hline &-1&-1&-6&-7&\color{blue}{-7}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrrr}1&-1&0&-5&-1&0&\color{orangered}{ 2 }\\& & -1& -1& -6& -7& \color{orangered}{-7} \\ \hline &\color{blue}{-1}&\color{blue}{-1}&\color{blue}{-6}&\color{blue}{-7}&\color{blue}{-7}&\color{orangered}{-5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -x^{4}-x^{3}-6x^{2}-7x-7 } $ with a remainder of $ \color{red}{ -5 } $.