Divide using synthetic division $$ ( 4x^{4}-8x^{3}+12x^{2}-6x+12 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&4&-8&12&-6&12\\& & 8& 0& 24& \color{black}{36} \\ \hline &\color{blue}{4}&\color{blue}{0}&\color{blue}{12}&\color{blue}{18}&\color{orangered}{48} \end{array} $$The solution is:
$$ \frac{ 4x^{4}-8x^{3}+12x^{2}-6x+12 }{ x-2 } = \color{blue}{4x^{3}+12x+18} ~+~ \frac{ \color{red}{ 48 } }{ x-2 } $$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}{ 8 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}2&4&\color{orangered}{ -8 }&12&-6&12\\& & \color{orangered}{8} & & & \\ \hline &4&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&-8&12&-6&12\\& & 8& \color{blue}{0} & & \\ \hline &4&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 0 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}2&4&-8&\color{orangered}{ 12 }&-6&12\\& & 8& \color{orangered}{0} & & \\ \hline &4&0&\color{orangered}{12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 12 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&-8&12&-6&12\\& & 8& 0& \color{blue}{24} & \\ \hline &4&0&\color{blue}{12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 24 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrrr}2&4&-8&12&\color{orangered}{ -6 }&12\\& & 8& 0& \color{orangered}{24} & \\ \hline &4&0&12&\color{orangered}{18}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 18 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&4&-8&12&-6&12\\& & 8& 0& 24& \color{blue}{36} \\ \hline &4&0&12&\color{blue}{18}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 36 } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrrrr}2&4&-8&12&-6&\color{orangered}{ 12 }\\& & 8& 0& 24& \color{orangered}{36} \\ \hline &\color{blue}{4}&\color{blue}{0}&\color{blue}{12}&\color{blue}{18}&\color{orangered}{48} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}+12x+18 } $ with a remainder of $ \color{red}{ 48 } $.