Divide using synthetic division $$ ( x^{5}-2x^{4}-4x^{3}-5x+6 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-2&1&-2&-4&0&-5&6\\& & -2& 8& -8& 16& \color{black}{-22} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{4}&\color{blue}{-8}&\color{blue}{11}&\color{orangered}{-16} \end{array} $$The solution is:
$$ \frac{ x^{5}-2x^{4}-4x^{3}-5x+6 }{ x+2 } = \color{blue}{x^{4}-4x^{3}+4x^{2}-8x+11} \color{red}{~-~} \frac{ \color{red}{ 16 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&-2&-4&0&-5&6\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-2&\color{orangered}{ 1 }&-2&-4&0&-5&6\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&-2&-4&0&-5&6\\& & \color{blue}{-2} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrrr}-2&1&\color{orangered}{ -2 }&-4&0&-5&6\\& & \color{orangered}{-2} & & & & \\ \hline &1&\color{orangered}{-4}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&-2&-4&0&-5&6\\& & -2& \color{blue}{8} & & & \\ \hline &1&\color{blue}{-4}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 8 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}-2&1&-2&\color{orangered}{ -4 }&0&-5&6\\& & -2& \color{orangered}{8} & & & \\ \hline &1&-4&\color{orangered}{4}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&-2&-4&0&-5&6\\& & -2& 8& \color{blue}{-8} & & \\ \hline &1&-4&\color{blue}{4}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrrr}-2&1&-2&-4&\color{orangered}{ 0 }&-5&6\\& & -2& 8& \color{orangered}{-8} & & \\ \hline &1&-4&4&\color{orangered}{-8}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&-2&-4&0&-5&6\\& & -2& 8& -8& \color{blue}{16} & \\ \hline &1&-4&4&\color{blue}{-8}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 16 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrrr}-2&1&-2&-4&0&\color{orangered}{ -5 }&6\\& & -2& 8& -8& \color{orangered}{16} & \\ \hline &1&-4&4&-8&\color{orangered}{11}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 11 } = \color{blue}{ -22 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&-2&-4&0&-5&6\\& & -2& 8& -8& 16& \color{blue}{-22} \\ \hline &1&-4&4&-8&\color{blue}{11}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -22 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrrr}-2&1&-2&-4&0&-5&\color{orangered}{ 6 }\\& & -2& 8& -8& 16& \color{orangered}{-22} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{4}&\color{blue}{-8}&\color{blue}{11}&\color{orangered}{-16} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-4x^{3}+4x^{2}-8x+11 } $ with a remainder of $ \color{red}{ -16 } $.