Divide using synthetic division $$ ( x^{4}+4x^{2}-6x+1 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&0&4&-6&1\\& & 2& 4& 16& \color{black}{20} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{8}&\color{blue}{10}&\color{orangered}{21} \end{array} $$The solution is:
$$ \frac{ x^{4}+4x^{2}-6x+1 }{ x-2 } = \color{blue}{x^{3}+2x^{2}+8x+10} ~+~ \frac{ \color{red}{ 21 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&0&4&-6&1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&0&4&-6&1\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&0&4&-6&1\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 2 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ 0 }&4&-6&1\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&0&4&-6&1\\& & 2& \color{blue}{4} & & \\ \hline &1&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 4 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}2&1&0&\color{orangered}{ 4 }&-6&1\\& & 2& \color{orangered}{4} & & \\ \hline &1&2&\color{orangered}{8}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 8 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&0&4&-6&1\\& & 2& 4& \color{blue}{16} & \\ \hline &1&2&\color{blue}{8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 16 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}2&1&0&4&\color{orangered}{ -6 }&1\\& & 2& 4& \color{orangered}{16} & \\ \hline &1&2&8&\color{orangered}{10}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 10 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&0&4&-6&1\\& & 2& 4& 16& \color{blue}{20} \\ \hline &1&2&8&\color{blue}{10}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 20 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrrr}2&1&0&4&-6&\color{orangered}{ 1 }\\& & 2& 4& 16& \color{orangered}{20} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{8}&\color{blue}{10}&\color{orangered}{21} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+2x^{2}+8x+10 } $ with a remainder of $ \color{red}{ 21 } $.