Divide using synthetic division $$ ( 4x^{4}-8x^{3}+12x^{2}-6x+12 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&4&-8&12&-6&12\\& & -8& 32& -88& \color{black}{188} \\ \hline &\color{blue}{4}&\color{blue}{-16}&\color{blue}{44}&\color{blue}{-94}&\color{orangered}{200} \end{array} $$The solution is:
$$ \frac{ 4x^{4}-8x^{3}+12x^{2}-6x+12 }{ x+2 } = \color{blue}{4x^{3}-16x^{2}+44x-94} ~+~ \frac{ \color{red}{ 200 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&-8&12&-6&12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 4 }&-8&12&-6&12\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&-8&12&-6&12\\& & \color{blue}{-8} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrr}-2&4&\color{orangered}{ -8 }&12&-6&12\\& & \color{orangered}{-8} & & & \\ \hline &4&\color{orangered}{-16}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&-8&12&-6&12\\& & -8& \color{blue}{32} & & \\ \hline &4&\color{blue}{-16}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 32 } = \color{orangered}{ 44 } $
$$ \begin{array}{c|rrrrr}-2&4&-8&\color{orangered}{ 12 }&-6&12\\& & -8& \color{orangered}{32} & & \\ \hline &4&-16&\color{orangered}{44}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 44 } = \color{blue}{ -88 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&-8&12&-6&12\\& & -8& 32& \color{blue}{-88} & \\ \hline &4&-16&\color{blue}{44}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -88 \right) } = \color{orangered}{ -94 } $
$$ \begin{array}{c|rrrrr}-2&4&-8&12&\color{orangered}{ -6 }&12\\& & -8& 32& \color{orangered}{-88} & \\ \hline &4&-16&44&\color{orangered}{-94}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -94 \right) } = \color{blue}{ 188 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&4&-8&12&-6&12\\& & -8& 32& -88& \color{blue}{188} \\ \hline &4&-16&44&\color{blue}{-94}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 188 } = \color{orangered}{ 200 } $
$$ \begin{array}{c|rrrrr}-2&4&-8&12&-6&\color{orangered}{ 12 }\\& & -8& 32& -88& \color{orangered}{188} \\ \hline &\color{blue}{4}&\color{blue}{-16}&\color{blue}{44}&\color{blue}{-94}&\color{orangered}{200} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-16x^{2}+44x-94 } $ with a remainder of $ \color{red}{ 200 } $.