Check if $ x-1 $ is a factor of $ 12x^{4}+5x^{3}-50x^{2}-20x+8 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&12&5&-50&-20&8\\& & 12& 17& -33& \color{black}{-53} \\ \hline &\color{blue}{12}&\color{blue}{17}&\color{blue}{-33}&\color{blue}{-53}&\color{orangered}{-45} \end{array} $$Because the remainder $ \left( \color{red}{ -45 } \right) $ is not zero, we conclude that the $ x-1 $ 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 -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&12&5&-50&-20&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 12 }&5&-50&-20&8\\& & & & & \\ \hline &\color{orangered}{12}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 12 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&12&5&-50&-20&8\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{12}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 12 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrrr}1&12&\color{orangered}{ 5 }&-50&-20&8\\& & \color{orangered}{12} & & & \\ \hline &12&\color{orangered}{17}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 17 } = \color{blue}{ 17 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&12&5&-50&-20&8\\& & 12& \color{blue}{17} & & \\ \hline &12&\color{blue}{17}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -50 } + \color{orangered}{ 17 } = \color{orangered}{ -33 } $
$$ \begin{array}{c|rrrrr}1&12&5&\color{orangered}{ -50 }&-20&8\\& & 12& \color{orangered}{17} & & \\ \hline &12&17&\color{orangered}{-33}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -33 \right) } = \color{blue}{ -33 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&12&5&-50&-20&8\\& & 12& 17& \color{blue}{-33} & \\ \hline &12&17&\color{blue}{-33}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ \left( -33 \right) } = \color{orangered}{ -53 } $
$$ \begin{array}{c|rrrrr}1&12&5&-50&\color{orangered}{ -20 }&8\\& & 12& 17& \color{orangered}{-33} & \\ \hline &12&17&-33&\color{orangered}{-53}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -53 \right) } = \color{blue}{ -53 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&12&5&-50&-20&8\\& & 12& 17& -33& \color{blue}{-53} \\ \hline &12&17&-33&\color{blue}{-53}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -53 \right) } = \color{orangered}{ -45 } $
$$ \begin{array}{c|rrrrr}1&12&5&-50&-20&\color{orangered}{ 8 }\\& & 12& 17& -33& \color{orangered}{-53} \\ \hline &\color{blue}{12}&\color{blue}{17}&\color{blue}{-33}&\color{blue}{-53}&\color{orangered}{-45} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -45 }\right)$.