Divide using synthetic division $$ ( x^{5}-2x^{3}+x^{2}-4 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-2&1&0&-2&1&0&-4\\& & -2& 4& -4& 6& \color{black}{-12} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{2}&\color{blue}{-3}&\color{blue}{6}&\color{orangered}{-16} \end{array} $$The solution is:
$$ \frac{ x^{5}-2x^{3}+x^{2}-4 }{ x+2 } = \color{blue}{x^{4}-2x^{3}+2x^{2}-3x+6} \color{red}{~-~} \frac{ \color{red}{ 16 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&-2&1&0&-4\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-2&\color{orangered}{ 1 }&0&-2&1&0&-4\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&-2&1&0&-4\\& & \color{blue}{-2} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}-2&1&\color{orangered}{ 0 }&-2&1&0&-4\\& & \color{orangered}{-2} & & & & \\ \hline &1&\color{orangered}{-2}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&-2&1&0&-4\\& & -2& \color{blue}{4} & & & \\ \hline &1&\color{blue}{-2}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 4 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}-2&1&0&\color{orangered}{ -2 }&1&0&-4\\& & -2& \color{orangered}{4} & & & \\ \hline &1&-2&\color{orangered}{2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 2 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&-2&1&0&-4\\& & -2& 4& \color{blue}{-4} & & \\ \hline &1&-2&\color{blue}{2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrr}-2&1&0&-2&\color{orangered}{ 1 }&0&-4\\& & -2& 4& \color{orangered}{-4} & & \\ \hline &1&-2&2&\color{orangered}{-3}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&-2&1&0&-4\\& & -2& 4& -4& \color{blue}{6} & \\ \hline &1&-2&2&\color{blue}{-3}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}-2&1&0&-2&1&\color{orangered}{ 0 }&-4\\& & -2& 4& -4& \color{orangered}{6} & \\ \hline &1&-2&2&-3&\color{orangered}{6}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 6 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&-2&1&0&-4\\& & -2& 4& -4& 6& \color{blue}{-12} \\ \hline &1&-2&2&-3&\color{blue}{6}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrrr}-2&1&0&-2&1&0&\color{orangered}{ -4 }\\& & -2& 4& -4& 6& \color{orangered}{-12} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{2}&\color{blue}{-3}&\color{blue}{6}&\color{orangered}{-16} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-2x^{3}+2x^{2}-3x+6 } $ with a remainder of $ \color{red}{ -16 } $.