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