Divide using synthetic division $$ ( x^{5}+2x^{3}-4x-2 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-2&1&0&2&0&-4&-2\\& & -2& 4& -12& 24& \color{black}{-40} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{6}&\color{blue}{-12}&\color{blue}{20}&\color{orangered}{-42} \end{array} $$The solution is:
$$ \frac{ x^{5}+2x^{3}-4x-2 }{ x+2 } = \color{blue}{x^{4}-2x^{3}+6x^{2}-12x+20} \color{red}{~-~} \frac{ \color{red}{ 42 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&2&0&-4&-2\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-2&\color{orangered}{ 1 }&0&2&0&-4&-2\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&2&0&-4&-2\\& & \color{blue}{-2} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}-2&1&\color{orangered}{ 0 }&2&0&-4&-2\\& & \color{orangered}{-2} & & & & \\ \hline &1&\color{orangered}{-2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&2&0&-4&-2\\& & -2& \color{blue}{4} & & & \\ \hline &1&\color{blue}{-2}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 4 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}-2&1&0&\color{orangered}{ 2 }&0&-4&-2\\& & -2& \color{orangered}{4} & & & \\ \hline &1&-2&\color{orangered}{6}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 6 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&2&0&-4&-2\\& & -2& 4& \color{blue}{-12} & & \\ \hline &1&-2&\color{blue}{6}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrrr}-2&1&0&2&\color{orangered}{ 0 }&-4&-2\\& & -2& 4& \color{orangered}{-12} & & \\ \hline &1&-2&6&\color{orangered}{-12}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&2&0&-4&-2\\& & -2& 4& -12& \color{blue}{24} & \\ \hline &1&-2&6&\color{blue}{-12}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 24 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrrr}-2&1&0&2&0&\color{orangered}{ -4 }&-2\\& & -2& 4& -12& \color{orangered}{24} & \\ \hline &1&-2&6&-12&\color{orangered}{20}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 20 } = \color{blue}{ -40 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&2&0&-4&-2\\& & -2& 4& -12& 24& \color{blue}{-40} \\ \hline &1&-2&6&-12&\color{blue}{20}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -40 \right) } = \color{orangered}{ -42 } $
$$ \begin{array}{c|rrrrrr}-2&1&0&2&0&-4&\color{orangered}{ -2 }\\& & -2& 4& -12& 24& \color{orangered}{-40} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{6}&\color{blue}{-12}&\color{blue}{20}&\color{orangered}{-42} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-2x^{3}+6x^{2}-12x+20 } $ with a remainder of $ \color{red}{ -42 } $.