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