Check if $ x+2 $ is a factor of $ x^{3}+6x^{2}+12x+12 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&1&6&12&12\\& & -2& -8& \color{black}{-8} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{4}&\color{orangered}{4} \end{array} $$Because the remainder $ \left( \color{red}{ 4 } \right) $ is not zero, we conclude that the $ x+2 $ is not a factor of $ x^{3}+6x^{2}+12x+12$.
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 + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&6&12&12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 1 }&6&12&12\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&6&12&12\\& & \color{blue}{-2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-2&1&\color{orangered}{ 6 }&12&12\\& & \color{orangered}{-2} & & \\ \hline &1&\color{orangered}{4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&6&12&12\\& & -2& \color{blue}{-8} & \\ \hline &1&\color{blue}{4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-2&1&6&\color{orangered}{ 12 }&12\\& & -2& \color{orangered}{-8} & \\ \hline &1&4&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&6&12&12\\& & -2& -8& \color{blue}{-8} \\ \hline &1&4&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-2&1&6&12&\color{orangered}{ 12 }\\& & -2& -8& \color{orangered}{-8} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{4}&\color{orangered}{4} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 4 }\right)$.