Check if $ x-2 $ is a factor of $ x^{3}+6x^{2}+5x-12 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&1&6&5&-12\\& & 2& 16& \color{black}{42} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{21}&\color{orangered}{30} \end{array} $$Because the remainder $ \left( \color{red}{ 30 } \right) $ is not zero, we conclude that the $ x-2 $ is not a factor of $ x^{3}+6x^{2}+5x-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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&6&5&-12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 1 }&6&5&-12\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&6&5&-12\\& & \color{blue}{2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 2 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}2&1&\color{orangered}{ 6 }&5&-12\\& & \color{orangered}{2} & & \\ \hline &1&\color{orangered}{8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 8 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&6&5&-12\\& & 2& \color{blue}{16} & \\ \hline &1&\color{blue}{8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 16 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrr}2&1&6&\color{orangered}{ 5 }&-12\\& & 2& \color{orangered}{16} & \\ \hline &1&8&\color{orangered}{21}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 21 } = \color{blue}{ 42 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&6&5&-12\\& & 2& 16& \color{blue}{42} \\ \hline &1&8&\color{blue}{21}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 42 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrr}2&1&6&5&\color{orangered}{ -12 }\\& & 2& 16& \color{orangered}{42} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{21}&\color{orangered}{30} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 30 }\right)$.