Divide using synthetic division $$ ( 6x^{4}-24x^{3}-33x^{2}+20x+3 ) \div ( x-5 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&6&-24&-33&20&3\\& & 30& 30& -15& \color{black}{25} \\ \hline &\color{blue}{6}&\color{blue}{6}&\color{blue}{-3}&\color{blue}{5}&\color{orangered}{28} \end{array} $$The solution is:
$$ \dfrac{ 6x^{4}-24x^{3}-33x^{2}+20x+3 }{ x-5 } = \color{blue}{6x^{3}+6x^{2}-3x+5} ~+~ \dfrac{ \color{red}{ 28 } }{ x-5 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&6&-24&-33&20&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 6 }&-24&-33&20&3\\& & & & & \\ \hline &\color{orangered}{6}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 6 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&6&-24&-33&20&3\\& & \color{blue}{30} & & & \\ \hline &\color{blue}{6}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 30 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}5&6&\color{orangered}{ -24 }&-33&20&3\\& & \color{orangered}{30} & & & \\ \hline &6&\color{orangered}{6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 6 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&6&-24&-33&20&3\\& & 30& \color{blue}{30} & & \\ \hline &6&\color{blue}{6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -33 } + \color{orangered}{ 30 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}5&6&-24&\color{orangered}{ -33 }&20&3\\& & 30& \color{orangered}{30} & & \\ \hline &6&6&\color{orangered}{-3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&6&-24&-33&20&3\\& & 30& 30& \color{blue}{-15} & \\ \hline &6&6&\color{blue}{-3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}5&6&-24&-33&\color{orangered}{ 20 }&3\\& & 30& 30& \color{orangered}{-15} & \\ \hline &6&6&-3&\color{orangered}{5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 5 } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&6&-24&-33&20&3\\& & 30& 30& -15& \color{blue}{25} \\ \hline &6&6&-3&\color{blue}{5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 25 } = \color{orangered}{ 28 } $
$$ \begin{array}{c|rrrrr}5&6&-24&-33&20&\color{orangered}{ 3 }\\& & 30& 30& -15& \color{orangered}{25} \\ \hline &\color{blue}{6}&\color{blue}{6}&\color{blue}{-3}&\color{blue}{5}&\color{orangered}{28} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{3}+6x^{2}-3x+5 } $ with a remainder of $ \color{red}{ 28 } $.