Divide using synthetic division $$ ( 2x^{5}+3x^{4}-3x^{3}+2x^{2}-4x+1 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-2&2&3&-3&2&-4&1\\& & -4& 2& 2& -8& \color{black}{24} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{-1}&\color{blue}{4}&\color{blue}{-12}&\color{orangered}{25} \end{array} $$The solution is:
$$ \frac{ 2x^{5}+3x^{4}-3x^{3}+2x^{2}-4x+1 }{ x+2 } = \color{blue}{2x^{4}-x^{3}-x^{2}+4x-12} ~+~ \frac{ \color{red}{ 25 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&2&3&-3&2&-4&1\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-2&\color{orangered}{ 2 }&3&-3&2&-4&1\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&2&3&-3&2&-4&1\\& & \color{blue}{-4} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}-2&2&\color{orangered}{ 3 }&-3&2&-4&1\\& & \color{orangered}{-4} & & & & \\ \hline &2&\color{orangered}{-1}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&2&3&-3&2&-4&1\\& & -4& \color{blue}{2} & & & \\ \hline &2&\color{blue}{-1}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 2 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}-2&2&3&\color{orangered}{ -3 }&2&-4&1\\& & -4& \color{orangered}{2} & & & \\ \hline &2&-1&\color{orangered}{-1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&2&3&-3&2&-4&1\\& & -4& 2& \color{blue}{2} & & \\ \hline &2&-1&\color{blue}{-1}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 2 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}-2&2&3&-3&\color{orangered}{ 2 }&-4&1\\& & -4& 2& \color{orangered}{2} & & \\ \hline &2&-1&-1&\color{orangered}{4}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&2&3&-3&2&-4&1\\& & -4& 2& 2& \color{blue}{-8} & \\ \hline &2&-1&-1&\color{blue}{4}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrrr}-2&2&3&-3&2&\color{orangered}{ -4 }&1\\& & -4& 2& 2& \color{orangered}{-8} & \\ \hline &2&-1&-1&4&\color{orangered}{-12}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&2&3&-3&2&-4&1\\& & -4& 2& 2& -8& \color{blue}{24} \\ \hline &2&-1&-1&4&\color{blue}{-12}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 24 } = \color{orangered}{ 25 } $
$$ \begin{array}{c|rrrrrr}-2&2&3&-3&2&-4&\color{orangered}{ 1 }\\& & -4& 2& 2& -8& \color{orangered}{24} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{-1}&\color{blue}{4}&\color{blue}{-12}&\color{orangered}{25} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}-x^{3}-x^{2}+4x-12 } $ with a remainder of $ \color{red}{ 25 } $.