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

The synthetic division table is:

$$ \begin{array}{c|rrrr}0&1&-3&4&-12\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{4}&\color{orangered}{-12} \end{array} $$

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

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.

$$ \begin{array}{c|rrrr}\color{blue}{0}&1&-3&4&-12\\& & & & \\ \hline &&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}0&\color{orangered}{ 1 }&-3&4&-12\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$

Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 1 } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrr}\color{blue}{0}&1&-3&4&-12\\& & \color{blue}{0} & & \\ \hline &\color{blue}{1}&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}0&1&\color{orangered}{ -3 }&4&-12\\& & \color{orangered}{0} & & \\ \hline &1&\color{orangered}{-3}&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrr}\color{blue}{0}&1&-3&4&-12\\& & 0& \color{blue}{0} & \\ \hline &1&\color{blue}{-3}&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 0 } = \color{orangered}{ 4 } $

$$ \begin{array}{c|rrrr}0&1&-3&\color{orangered}{ 4 }&-12\\& & 0& \color{orangered}{0} & \\ \hline &1&-3&\color{orangered}{4}& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 4 } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrr}\color{blue}{0}&1&-3&4&-12\\& & 0& 0& \color{blue}{0} \\ \hline &1&-3&\color{blue}{4}& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 0 } = \color{orangered}{ -12 } $

$$ \begin{array}{c|rrrr}0&1&-3&4&\color{orangered}{ -12 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{4}&\color{orangered}{-12} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ -12 }\right)$.

Synthetic Division Calculator