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