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