Divide using synthetic division $$ ( x^{5}+2x^{4}-3x^{3}+x-1 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-2&1&2&-3&0&1&-1\\& & -2& 0& 6& -12& \color{black}{22} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-3}&\color{blue}{6}&\color{blue}{-11}&\color{orangered}{21} \end{array} $$The solution is:
$$ \frac{ x^{5}+2x^{4}-3x^{3}+x-1 }{ x+2 } = \color{blue}{x^{4}-3x^{2}+6x-11} ~+~ \frac{ \color{red}{ 21 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&2&-3&0&1&-1\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-2&\color{orangered}{ 1 }&2&-3&0&1&-1\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&2&-3&0&1&-1\\& & \color{blue}{-2} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-2&1&\color{orangered}{ 2 }&-3&0&1&-1\\& & \color{orangered}{-2} & & & & \\ \hline &1&\color{orangered}{0}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&2&-3&0&1&-1\\& & -2& \color{blue}{0} & & & \\ \hline &1&\color{blue}{0}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 0 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrr}-2&1&2&\color{orangered}{ -3 }&0&1&-1\\& & -2& \color{orangered}{0} & & & \\ \hline &1&0&\color{orangered}{-3}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&2&-3&0&1&-1\\& & -2& 0& \color{blue}{6} & & \\ \hline &1&0&\color{blue}{-3}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}-2&1&2&-3&\color{orangered}{ 0 }&1&-1\\& & -2& 0& \color{orangered}{6} & & \\ \hline &1&0&-3&\color{orangered}{6}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 6 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&2&-3&0&1&-1\\& & -2& 0& 6& \color{blue}{-12} & \\ \hline &1&0&-3&\color{blue}{6}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -11 } $
$$ \begin{array}{c|rrrrrr}-2&1&2&-3&0&\color{orangered}{ 1 }&-1\\& & -2& 0& 6& \color{orangered}{-12} & \\ \hline &1&0&-3&6&\color{orangered}{-11}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ 22 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&2&-3&0&1&-1\\& & -2& 0& 6& -12& \color{blue}{22} \\ \hline &1&0&-3&6&\color{blue}{-11}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 22 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrrrr}-2&1&2&-3&0&1&\color{orangered}{ -1 }\\& & -2& 0& 6& -12& \color{orangered}{22} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-3}&\color{blue}{6}&\color{blue}{-11}&\color{orangered}{21} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-3x^{2}+6x-11 } $ with a remainder of $ \color{red}{ 21 } $.