Check if $ x+4 $ is a factor of $ 3x^{3}-3x^{2}-108 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&3&-3&0&-108\\& & -12& 60& \color{black}{-240} \\ \hline &\color{blue}{3}&\color{blue}{-15}&\color{blue}{60}&\color{orangered}{-348} \end{array} $$Because the remainder $ \left( \color{red}{ -348 } \right) $ is not zero, we conclude that the $ x+4 $ 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 + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&3&-3&0&-108\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 3 }&-3&0&-108\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 3 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&3&-3&0&-108\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrr}-4&3&\color{orangered}{ -3 }&0&-108\\& & \color{orangered}{-12} & & \\ \hline &3&\color{orangered}{-15}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -15 \right) } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&3&-3&0&-108\\& & -12& \color{blue}{60} & \\ \hline &3&\color{blue}{-15}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 60 } = \color{orangered}{ 60 } $
$$ \begin{array}{c|rrrr}-4&3&-3&\color{orangered}{ 0 }&-108\\& & -12& \color{orangered}{60} & \\ \hline &3&-15&\color{orangered}{60}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 60 } = \color{blue}{ -240 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&3&-3&0&-108\\& & -12& 60& \color{blue}{-240} \\ \hline &3&-15&\color{blue}{60}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -108 } + \color{orangered}{ \left( -240 \right) } = \color{orangered}{ -348 } $
$$ \begin{array}{c|rrrr}-4&3&-3&0&\color{orangered}{ -108 }\\& & -12& 60& \color{orangered}{-240} \\ \hline &\color{blue}{3}&\color{blue}{-15}&\color{blue}{60}&\color{orangered}{-348} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -348 }\right)$.