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