Check if $ 4 $ is a factor of $ 12x^{4}+5x^{3}-50x^{2}-20x+8 $.

The synthetic division table is:

$$ \begin{array}{c|rrrrr}0&12&5&-50&-20&8\\& & 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{12}&\color{blue}{5}&\color{blue}{-50}&\color{blue}{-20}&\color{orangered}{8} \end{array} $$

Because the remainder $ \left( \color{red}{ 8 } \right) $ is not zero, we conclude that the $ x $ is not a factor of $ 12x^{4}+5x^{3}-50x^{2}-20x+8$.

First we need to create a synthetic division table.

Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.

$$ \begin{array}{c|rrrrr}\color{blue}{0}&12&5&-50&-20&8\\& & & & & \\ \hline &&&&& \end{array} $$

Step 1 : Bring down the leading coefficient to the bottom row.

$$ \begin{array}{c|rrrrr}0&\color{orangered}{ 12 }&5&-50&-20&8\\& & & & & \\ \hline &\color{orangered}{12}&&&& \end{array} $$

Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 12 } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{0}&12&5&-50&-20&8\\& & \color{blue}{0} & & & \\ \hline &\color{blue}{12}&&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 0 } = \color{orangered}{ 5 } $

$$ \begin{array}{c|rrrrr}0&12&\color{orangered}{ 5 }&-50&-20&8\\& & \color{orangered}{0} & & & \\ \hline &12&\color{orangered}{5}&&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 5 } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{0}&12&5&-50&-20&8\\& & 0& \color{blue}{0} & & \\ \hline &12&\color{blue}{5}&&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ -50 } + \color{orangered}{ 0 } = \color{orangered}{ -50 } $

$$ \begin{array}{c|rrrrr}0&12&5&\color{orangered}{ -50 }&-20&8\\& & 0& \color{orangered}{0} & & \\ \hline &12&5&\color{orangered}{-50}&& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -50 \right) } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{0}&12&5&-50&-20&8\\& & 0& 0& \color{blue}{0} & \\ \hline &12&5&\color{blue}{-50}&& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 0 } = \color{orangered}{ -20 } $

$$ \begin{array}{c|rrrrr}0&12&5&-50&\color{orangered}{ -20 }&8\\& & 0& 0& \color{orangered}{0} & \\ \hline &12&5&-50&\color{orangered}{-20}& \end{array} $$

Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{0}&12&5&-50&-20&8\\& & 0& 0& 0& \color{blue}{0} \\ \hline &12&5&-50&\color{blue}{-20}& \end{array} $$

Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 0 } = \color{orangered}{ 8 } $

$$ \begin{array}{c|rrrrr}0&12&5&-50&-20&\color{orangered}{ 8 }\\& & 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{12}&\color{blue}{5}&\color{blue}{-50}&\color{blue}{-20}&\color{orangered}{8} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ 8 }\right)$.

Synthetic Division Calculator