Divide using synthetic division $$ ( x^{4}-2x^{3}+x^{2}+12x-6 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&-2&1&12&-6\\& & 2& 0& 2& \color{black}{28} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{1}&\color{blue}{14}&\color{orangered}{22} \end{array} $$The solution is:
$$ \frac{ x^{4}-2x^{3}+x^{2}+12x-6 }{ x-2 } = \color{blue}{x^{3}+x+14} ~+~ \frac{ \color{red}{ 22 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-2&1&12&-6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&-2&1&12&-6\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-2&1&12&-6\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 2 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ -2 }&1&12&-6\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{0}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-2&1&12&-6\\& & 2& \color{blue}{0} & & \\ \hline &1&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 0 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}2&1&-2&\color{orangered}{ 1 }&12&-6\\& & 2& \color{orangered}{0} & & \\ \hline &1&0&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-2&1&12&-6\\& & 2& 0& \color{blue}{2} & \\ \hline &1&0&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 2 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrrr}2&1&-2&1&\color{orangered}{ 12 }&-6\\& & 2& 0& \color{orangered}{2} & \\ \hline &1&0&1&\color{orangered}{14}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 14 } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-2&1&12&-6\\& & 2& 0& 2& \color{blue}{28} \\ \hline &1&0&1&\color{blue}{14}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 28 } = \color{orangered}{ 22 } $
$$ \begin{array}{c|rrrrr}2&1&-2&1&12&\color{orangered}{ -6 }\\& & 2& 0& 2& \color{orangered}{28} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{1}&\color{blue}{14}&\color{orangered}{22} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+x+14 } $ with a remainder of $ \color{red}{ 22 } $.