The synthetic division table is:
$$ \begin{array}{c|rrrr}4&4&-21&22&-8\\& & 16& -20& \color{black}{8} \\ \hline &\color{blue}{4}&\color{blue}{-5}&\color{blue}{2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-21x^{2}+22x-8 }{ x-4 } = \color{blue}{4x^{2}-5x+2} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&-21&22&-8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 4 }&-21&22&-8\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 4 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&-21&22&-8\\& & \color{blue}{16} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 16 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}4&4&\color{orangered}{ -21 }&22&-8\\& & \color{orangered}{16} & & \\ \hline &4&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&-21&22&-8\\& & 16& \color{blue}{-20} & \\ \hline &4&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 22 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}4&4&-21&\color{orangered}{ 22 }&-8\\& & 16& \color{orangered}{-20} & \\ \hline &4&-5&\color{orangered}{2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 2 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&-21&22&-8\\& & 16& -20& \color{blue}{8} \\ \hline &4&-5&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 8 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}4&4&-21&22&\color{orangered}{ -8 }\\& & 16& -20& \color{orangered}{8} \\ \hline &\color{blue}{4}&\color{blue}{-5}&\color{blue}{2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-5x+2 } $ with a remainder of $ \color{red}{ 0 } $.