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