Divide using synthetic division $$ ( x^{4}+11x^{3}+20x^{2}+9x-1 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&1&11&20&9&-1\\& & -1& -10& -10& \color{black}{1} \\ \hline &\color{blue}{1}&\color{blue}{10}&\color{blue}{10}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}+11x^{3}+20x^{2}+9x-1 }{ x+1 } = \color{blue}{x^{3}+10x^{2}+10x-1} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&11&20&9&-1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 1 }&11&20&9&-1\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 1 } = \color{blue}{ -1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&11&20&9&-1\\& & \color{blue}{-1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}-1&1&\color{orangered}{ 11 }&20&9&-1\\& & \color{orangered}{-1} & & & \\ \hline &1&\color{orangered}{10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 10 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&11&20&9&-1\\& & -1& \color{blue}{-10} & & \\ \hline &1&\color{blue}{10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}-1&1&11&\color{orangered}{ 20 }&9&-1\\& & -1& \color{orangered}{-10} & & \\ \hline &1&10&\color{orangered}{10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 10 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&11&20&9&-1\\& & -1& -10& \color{blue}{-10} & \\ \hline &1&10&\color{blue}{10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-1&1&11&20&\color{orangered}{ 9 }&-1\\& & -1& -10& \color{orangered}{-10} & \\ \hline &1&10&10&\color{orangered}{-1}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&11&20&9&-1\\& & -1& -10& -10& \color{blue}{1} \\ \hline &1&10&10&\color{blue}{-1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 1 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-1&1&11&20&9&\color{orangered}{ -1 }\\& & -1& -10& -10& \color{orangered}{1} \\ \hline &\color{blue}{1}&\color{blue}{10}&\color{blue}{10}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+10x^{2}+10x-1 } $ with a remainder of $ \color{red}{ 0 } $.