Divide using synthetic division $$ ( x^{5}+6x^{4}+x^{3}+11x^{2}+25x+20 ) \div ( x+1 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-1&1&6&1&11&25&20\\& & -1& -5& 4& -15& \color{black}{-10} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-4}&\color{blue}{15}&\color{blue}{10}&\color{orangered}{10} \end{array} $$The solution is:
$$ \dfrac{ x^{5}+6x^{4}+x^{3}+11x^{2}+25x+20 }{ x+1 } = \color{blue}{x^{4}+5x^{3}-4x^{2}+15x+10} ~+~ \dfrac{ \color{red}{ 10 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&1&6&1&11&25&20\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-1&\color{orangered}{ 1 }&6&1&11&25&20\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 1 } = \color{blue}{ -1 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&1&6&1&11&25&20\\& & \color{blue}{-1} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}-1&1&\color{orangered}{ 6 }&1&11&25&20\\& & \color{orangered}{-1} & & & & \\ \hline &1&\color{orangered}{5}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&1&6&1&11&25&20\\& & -1& \color{blue}{-5} & & & \\ \hline &1&\color{blue}{5}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrrr}-1&1&6&\color{orangered}{ 1 }&11&25&20\\& & -1& \color{orangered}{-5} & & & \\ \hline &1&5&\color{orangered}{-4}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&1&6&1&11&25&20\\& & -1& -5& \color{blue}{4} & & \\ \hline &1&5&\color{blue}{-4}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ 4 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrrr}-1&1&6&1&\color{orangered}{ 11 }&25&20\\& & -1& -5& \color{orangered}{4} & & \\ \hline &1&5&-4&\color{orangered}{15}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 15 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&1&6&1&11&25&20\\& & -1& -5& 4& \color{blue}{-15} & \\ \hline &1&5&-4&\color{blue}{15}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrr}-1&1&6&1&11&\color{orangered}{ 25 }&20\\& & -1& -5& 4& \color{orangered}{-15} & \\ \hline &1&5&-4&15&\color{orangered}{10}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 10 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&1&6&1&11&25&20\\& & -1& -5& 4& -15& \color{blue}{-10} \\ \hline &1&5&-4&15&\color{blue}{10}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrr}-1&1&6&1&11&25&\color{orangered}{ 20 }\\& & -1& -5& 4& -15& \color{orangered}{-10} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-4}&\color{blue}{15}&\color{blue}{10}&\color{orangered}{10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+5x^{3}-4x^{2}+15x+10 } $ with a remainder of $ \color{red}{ 10 } $.