Divide using synthetic division $$ ( 6x^{5}-x^{4}-20x^{2}+5x+8 ) \div ( 2x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&6&-1&0&-20&5&8\\& & 18& 51& 153& 399& \color{black}{1212} \\ \hline &\color{blue}{6}&\color{blue}{17}&\color{blue}{51}&\color{blue}{133}&\color{blue}{404}&\color{orangered}{1220} \end{array} $$The solution is:
$$ \frac{ 6x^{5}-x^{4}-20x^{2}+5x+8 }{ x-3 } = \color{blue}{6x^{4}+17x^{3}+51x^{2}+133x+404} ~+~ \frac{ \color{red}{ 1220 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&6&-1&0&-20&5&8\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 6 }&-1&0&-20&5&8\\& & & & & & \\ \hline &\color{orangered}{6}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 6 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&6&-1&0&-20&5&8\\& & \color{blue}{18} & & & & \\ \hline &\color{blue}{6}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 18 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrrrr}3&6&\color{orangered}{ -1 }&0&-20&5&8\\& & \color{orangered}{18} & & & & \\ \hline &6&\color{orangered}{17}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 17 } = \color{blue}{ 51 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&6&-1&0&-20&5&8\\& & 18& \color{blue}{51} & & & \\ \hline &6&\color{blue}{17}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 51 } = \color{orangered}{ 51 } $
$$ \begin{array}{c|rrrrrr}3&6&-1&\color{orangered}{ 0 }&-20&5&8\\& & 18& \color{orangered}{51} & & & \\ \hline &6&17&\color{orangered}{51}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 51 } = \color{blue}{ 153 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&6&-1&0&-20&5&8\\& & 18& 51& \color{blue}{153} & & \\ \hline &6&17&\color{blue}{51}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 153 } = \color{orangered}{ 133 } $
$$ \begin{array}{c|rrrrrr}3&6&-1&0&\color{orangered}{ -20 }&5&8\\& & 18& 51& \color{orangered}{153} & & \\ \hline &6&17&51&\color{orangered}{133}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 133 } = \color{blue}{ 399 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&6&-1&0&-20&5&8\\& & 18& 51& 153& \color{blue}{399} & \\ \hline &6&17&51&\color{blue}{133}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 399 } = \color{orangered}{ 404 } $
$$ \begin{array}{c|rrrrrr}3&6&-1&0&-20&\color{orangered}{ 5 }&8\\& & 18& 51& 153& \color{orangered}{399} & \\ \hline &6&17&51&133&\color{orangered}{404}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 404 } = \color{blue}{ 1212 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&6&-1&0&-20&5&8\\& & 18& 51& 153& 399& \color{blue}{1212} \\ \hline &6&17&51&133&\color{blue}{404}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 1212 } = \color{orangered}{ 1220 } $
$$ \begin{array}{c|rrrrrr}3&6&-1&0&-20&5&\color{orangered}{ 8 }\\& & 18& 51& 153& 399& \color{orangered}{1212} \\ \hline &\color{blue}{6}&\color{blue}{17}&\color{blue}{51}&\color{blue}{133}&\color{blue}{404}&\color{orangered}{1220} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{4}+17x^{3}+51x^{2}+133x+404 } $ with a remainder of $ \color{red}{ 1220 } $.