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