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