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