Check if $ x-6 $ is a factor of $ 3x^{3}-3x^{2}-108 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}6&3&-3&0&-108\\& & 18& 90& \color{black}{540} \\ \hline &\color{blue}{3}&\color{blue}{15}&\color{blue}{90}&\color{orangered}{432} \end{array} $$Because the remainder $ \left( \color{red}{ 432 } \right) $ is not zero, we conclude that the $ x-6 $ is not a factor of $ 3x^{3}-3x^{2}-108$.
Explanation
First we need to create a synthetic division table.
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -6 = 0 $ ( $ x = \color{blue}{ 6 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{6}&3&-3&0&-108\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}6&\color{orangered}{ 3 }&-3&0&-108\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 3 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&3&-3&0&-108\\& & \color{blue}{18} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 18 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}6&3&\color{orangered}{ -3 }&0&-108\\& & \color{orangered}{18} & & \\ \hline &3&\color{orangered}{15}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 15 } = \color{blue}{ 90 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&3&-3&0&-108\\& & 18& \color{blue}{90} & \\ \hline &3&\color{blue}{15}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 90 } = \color{orangered}{ 90 } $
$$ \begin{array}{c|rrrr}6&3&-3&\color{orangered}{ 0 }&-108\\& & 18& \color{orangered}{90} & \\ \hline &3&15&\color{orangered}{90}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 90 } = \color{blue}{ 540 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&3&-3&0&-108\\& & 18& 90& \color{blue}{540} \\ \hline &3&15&\color{blue}{90}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -108 } + \color{orangered}{ 540 } = \color{orangered}{ 432 } $
$$ \begin{array}{c|rrrr}6&3&-3&0&\color{orangered}{ -108 }\\& & 18& 90& \color{orangered}{540} \\ \hline &\color{blue}{3}&\color{blue}{15}&\color{blue}{90}&\color{orangered}{432} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 432 }\right)$.