The synthetic division table is:
$$ \begin{array}{c|rrrr}2&1&-12&-5&50\\& & 2& -20& \color{black}{-50} \\ \hline &\color{blue}{1}&\color{blue}{-10}&\color{blue}{-25}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}-12x^{2}-5x+50 }{ x-2 } = \color{blue}{x^{2}-10x-25} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-12&-5&50\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 1 }&-12&-5&50\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-12&-5&50\\& & \color{blue}{2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 2 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}2&1&\color{orangered}{ -12 }&-5&50\\& & \color{orangered}{2} & & \\ \hline &1&\color{orangered}{-10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-12&-5&50\\& & 2& \color{blue}{-20} & \\ \hline &1&\color{blue}{-10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -25 } $
$$ \begin{array}{c|rrrr}2&1&-12&\color{orangered}{ -5 }&50\\& & 2& \color{orangered}{-20} & \\ \hline &1&-10&\color{orangered}{-25}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -25 \right) } = \color{blue}{ -50 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-12&-5&50\\& & 2& -20& \color{blue}{-50} \\ \hline &1&-10&\color{blue}{-25}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 50 } + \color{orangered}{ \left( -50 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}2&1&-12&-5&\color{orangered}{ 50 }\\& & 2& -20& \color{orangered}{-50} \\ \hline &\color{blue}{1}&\color{blue}{-10}&\color{blue}{-25}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-10x-25 } $ with a remainder of $ \color{red}{ 0 } $.