Divide using synthetic division $$ ( 5x^{4}-17x^{3}-18x^{2}+19x+20 ) \div ( x-4 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&5&-17&-18&19&20\\& & 20& 12& -24& \color{black}{-20} \\ \hline &\color{blue}{5}&\color{blue}{3}&\color{blue}{-6}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$The solution is:
$$ \dfrac{ 5x^{4}-17x^{3}-18x^{2}+19x+20 }{ x-4 } = \color{blue}{5x^{3}+3x^{2}-6x-5} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&-17&-18&19&20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 5 }&-17&-18&19&20\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 5 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&-17&-18&19&20\\& & \color{blue}{20} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 20 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}4&5&\color{orangered}{ -17 }&-18&19&20\\& & \color{orangered}{20} & & & \\ \hline &5&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 3 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&-17&-18&19&20\\& & 20& \color{blue}{12} & & \\ \hline &5&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 12 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}4&5&-17&\color{orangered}{ -18 }&19&20\\& & 20& \color{orangered}{12} & & \\ \hline &5&3&\color{orangered}{-6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&-17&-18&19&20\\& & 20& 12& \color{blue}{-24} & \\ \hline &5&3&\color{blue}{-6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}4&5&-17&-18&\color{orangered}{ 19 }&20\\& & 20& 12& \color{orangered}{-24} & \\ \hline &5&3&-6&\color{orangered}{-5}& \end{array} $$Step 8 : 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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&5&-17&-18&19&20\\& & 20& 12& -24& \color{blue}{-20} \\ \hline &5&3&-6&\color{blue}{-5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}4&5&-17&-18&19&\color{orangered}{ 20 }\\& & 20& 12& -24& \color{orangered}{-20} \\ \hline &\color{blue}{5}&\color{blue}{3}&\color{blue}{-6}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}+3x^{2}-6x-5 } $ with a remainder of $ \color{red}{ 0 } $.