Check if $ x-2 $ is a factor of $ 12x^{4}+5x^{3}-50x^{2}-20x+8 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&12&5&-50&-20&8\\& & 24& 58& 16& \color{black}{-8} \\ \hline &\color{blue}{12}&\color{blue}{29}&\color{blue}{8}&\color{blue}{-4}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x-2 $ is a factor of the $ 12x^{4}+5x^{3}-50x^{2}-20x+8 $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&12&5&-50&-20&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 12 }&5&-50&-20&8\\& & & & & \\ \hline &\color{orangered}{12}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 12 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&12&5&-50&-20&8\\& & \color{blue}{24} & & & \\ \hline &\color{blue}{12}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 24 } = \color{orangered}{ 29 } $
$$ \begin{array}{c|rrrrr}2&12&\color{orangered}{ 5 }&-50&-20&8\\& & \color{orangered}{24} & & & \\ \hline &12&\color{orangered}{29}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 29 } = \color{blue}{ 58 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&12&5&-50&-20&8\\& & 24& \color{blue}{58} & & \\ \hline &12&\color{blue}{29}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -50 } + \color{orangered}{ 58 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}2&12&5&\color{orangered}{ -50 }&-20&8\\& & 24& \color{orangered}{58} & & \\ \hline &12&29&\color{orangered}{8}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 8 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&12&5&-50&-20&8\\& & 24& 58& \color{blue}{16} & \\ \hline &12&29&\color{blue}{8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 16 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}2&12&5&-50&\color{orangered}{ -20 }&8\\& & 24& 58& \color{orangered}{16} & \\ \hline &12&29&8&\color{orangered}{-4}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&12&5&-50&-20&8\\& & 24& 58& 16& \color{blue}{-8} \\ \hline &12&29&8&\color{blue}{-4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}2&12&5&-50&-20&\color{orangered}{ 8 }\\& & 24& 58& 16& \color{orangered}{-8} \\ \hline &\color{blue}{12}&\color{blue}{29}&\color{blue}{8}&\color{blue}{-4}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.