The synthetic division table is:
$$ \begin{array}{c|rrrr}4&3&0&-5&2\\& & 12& 48& \color{black}{172} \\ \hline &\color{blue}{3}&\color{blue}{12}&\color{blue}{43}&\color{orangered}{174} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-5x+2 }{ x-4 } = \color{blue}{3x^{2}+12x+43} ~+~ \frac{ \color{red}{ 174 } }{ x-4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&0&-5&2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 3 }&0&-5&2\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 3 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&0&-5&2\\& & \color{blue}{12} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}4&3&\color{orangered}{ 0 }&-5&2\\& & \color{orangered}{12} & & \\ \hline &3&\color{orangered}{12}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 12 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&0&-5&2\\& & 12& \color{blue}{48} & \\ \hline &3&\color{blue}{12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 48 } = \color{orangered}{ 43 } $
$$ \begin{array}{c|rrrr}4&3&0&\color{orangered}{ -5 }&2\\& & 12& \color{orangered}{48} & \\ \hline &3&12&\color{orangered}{43}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 43 } = \color{blue}{ 172 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&0&-5&2\\& & 12& 48& \color{blue}{172} \\ \hline &3&12&\color{blue}{43}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 172 } = \color{orangered}{ 174 } $
$$ \begin{array}{c|rrrr}4&3&0&-5&\color{orangered}{ 2 }\\& & 12& 48& \color{orangered}{172} \\ \hline &\color{blue}{3}&\color{blue}{12}&\color{blue}{43}&\color{orangered}{174} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+12x+43 } $ with a remainder of $ \color{red}{ 174 } $.