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