Divide using synthetic division $$ ( 2x^{4}+3x^{2}-5x-2 ) \div ( x+2 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&2&0&3&-5&-2\\& & -4& 8& -22& \color{black}{54} \\ \hline &\color{blue}{2}&\color{blue}{-4}&\color{blue}{11}&\color{blue}{-27}&\color{orangered}{52} \end{array} $$The solution is:
$$ \dfrac{ 2x^{4}+3x^{2}-5x-2 }{ x+2 } = \color{blue}{2x^{3}-4x^{2}+11x-27} ~+~ \dfrac{ \color{red}{ 52 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&0&3&-5&-2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 2 }&0&3&-5&-2\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&0&3&-5&-2\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-2&2&\color{orangered}{ 0 }&3&-5&-2\\& & \color{orangered}{-4} & & & \\ \hline &2&\color{orangered}{-4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&0&3&-5&-2\\& & -4& \color{blue}{8} & & \\ \hline &2&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 8 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrr}-2&2&0&\color{orangered}{ 3 }&-5&-2\\& & -4& \color{orangered}{8} & & \\ \hline &2&-4&\color{orangered}{11}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 11 } = \color{blue}{ -22 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&0&3&-5&-2\\& & -4& 8& \color{blue}{-22} & \\ \hline &2&-4&\color{blue}{11}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -22 \right) } = \color{orangered}{ -27 } $
$$ \begin{array}{c|rrrrr}-2&2&0&3&\color{orangered}{ -5 }&-2\\& & -4& 8& \color{orangered}{-22} & \\ \hline &2&-4&11&\color{orangered}{-27}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -27 \right) } = \color{blue}{ 54 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&0&3&-5&-2\\& & -4& 8& -22& \color{blue}{54} \\ \hline &2&-4&11&\color{blue}{-27}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 54 } = \color{orangered}{ 52 } $
$$ \begin{array}{c|rrrrr}-2&2&0&3&-5&\color{orangered}{ -2 }\\& & -4& 8& -22& \color{orangered}{54} \\ \hline &\color{blue}{2}&\color{blue}{-4}&\color{blue}{11}&\color{blue}{-27}&\color{orangered}{52} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-4x^{2}+11x-27 } $ with a remainder of $ \color{red}{ 52 } $.