The synthetic division table is:
$$ \begin{array}{c|rrr}0&1&-3&-33\\& & 0& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{orangered}{-33} \end{array} $$Because the remainder $ \left( \color{red}{ -33 } \right) $ is not zero, we conclude that the $ x $ is not a factor of $ x^{2}-3x-33$.
First we need to create a synthetic division table.
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrr}\color{blue}{0}&1&-3&-33\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}0&\color{orangered}{ 1 }&-3&-33\\& & & \\ \hline &\color{orangered}{1}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 1 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrr}\color{blue}{0}&1&-3&-33\\& & \color{blue}{0} & \\ \hline &\color{blue}{1}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 0 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrr}0&1&\color{orangered}{ -3 }&-33\\& & \color{orangered}{0} & \\ \hline &1&\color{orangered}{-3}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrr}\color{blue}{0}&1&-3&-33\\& & 0& \color{blue}{0} \\ \hline &1&\color{blue}{-3}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -33 } + \color{orangered}{ 0 } = \color{orangered}{ -33 } $
$$ \begin{array}{c|rrr}0&1&-3&\color{orangered}{ -33 }\\& & 0& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{orangered}{-33} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -33 }\right)$.