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