Check if $ x-4 $ is a factor of $ 12x^{4}+5x^{3}-50x^{2}-20x+8 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&12&5&-50&-20&8\\& & 48& 212& 648& \color{black}{2512} \\ \hline &\color{blue}{12}&\color{blue}{53}&\color{blue}{162}&\color{blue}{628}&\color{orangered}{2520} \end{array} $$Because the remainder $ \left( \color{red}{ 2520 } \right) $ is not zero, we conclude that the $ x-4 $ is not a factor of $ 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 -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&12&5&-50&-20&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 12 }&5&-50&-20&8\\& & & & & \\ \hline &\color{orangered}{12}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 12 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&12&5&-50&-20&8\\& & \color{blue}{48} & & & \\ \hline &\color{blue}{12}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 48 } = \color{orangered}{ 53 } $
$$ \begin{array}{c|rrrrr}4&12&\color{orangered}{ 5 }&-50&-20&8\\& & \color{orangered}{48} & & & \\ \hline &12&\color{orangered}{53}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 53 } = \color{blue}{ 212 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&12&5&-50&-20&8\\& & 48& \color{blue}{212} & & \\ \hline &12&\color{blue}{53}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -50 } + \color{orangered}{ 212 } = \color{orangered}{ 162 } $
$$ \begin{array}{c|rrrrr}4&12&5&\color{orangered}{ -50 }&-20&8\\& & 48& \color{orangered}{212} & & \\ \hline &12&53&\color{orangered}{162}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 162 } = \color{blue}{ 648 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&12&5&-50&-20&8\\& & 48& 212& \color{blue}{648} & \\ \hline &12&53&\color{blue}{162}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 648 } = \color{orangered}{ 628 } $
$$ \begin{array}{c|rrrrr}4&12&5&-50&\color{orangered}{ -20 }&8\\& & 48& 212& \color{orangered}{648} & \\ \hline &12&53&162&\color{orangered}{628}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 628 } = \color{blue}{ 2512 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&12&5&-50&-20&8\\& & 48& 212& 648& \color{blue}{2512} \\ \hline &12&53&162&\color{blue}{628}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 2512 } = \color{orangered}{ 2520 } $
$$ \begin{array}{c|rrrrr}4&12&5&-50&-20&\color{orangered}{ 8 }\\& & 48& 212& 648& \color{orangered}{2512} \\ \hline &\color{blue}{12}&\color{blue}{53}&\color{blue}{162}&\color{blue}{628}&\color{orangered}{2520} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 2520 }\right)$.