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