Check if $ 1 $ is a factor of $ 3x^{4}-16x^{3}+21x^{2}+4x-12 $.

The synthetic division table is:

$$ \begin{array}{c|rrrrr}0&3&-16&21&4&-12\\& & 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{3}&\color{blue}{-16}&\color{blue}{21}&\color{blue}{4}&\color{orangered}{-12} \end{array} $$

Because the remainder $ \left( \color{red}{ -12 } \right) $ is not zero, we conclude that the $ x $ is not a factor of $ 3x^{4}-16x^{3}+21x^{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.

$$ \begin{array}{c|rrrrr}\color{blue}{0}&3&-16&21&4&-12\\& & & & & \\ \hline &&&&& \end{array} $$

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

$$ \begin{array}{c|rrrrr}0&\color{orangered}{ 3 }&-16&21&4&-12\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$

Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 3 } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{0}&3&-16&21&4&-12\\& & \color{blue}{0} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 0 } = \color{orangered}{ -16 } $

$$ \begin{array}{c|rrrrr}0&3&\color{orangered}{ -16 }&21&4&-12\\& & \color{orangered}{0} & & & \\ \hline &3&\color{orangered}{-16}&&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{0}&3&-16&21&4&-12\\& & 0& \color{blue}{0} & & \\ \hline &3&\color{blue}{-16}&&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ 0 } = \color{orangered}{ 21 } $

$$ \begin{array}{c|rrrrr}0&3&-16&\color{orangered}{ 21 }&4&-12\\& & 0& \color{orangered}{0} & & \\ \hline &3&-16&\color{orangered}{21}&& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 21 } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{0}&3&-16&21&4&-12\\& & 0& 0& \color{blue}{0} & \\ \hline &3&-16&\color{blue}{21}&& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 0 } = \color{orangered}{ 4 } $

$$ \begin{array}{c|rrrrr}0&3&-16&21&\color{orangered}{ 4 }&-12\\& & 0& 0& \color{orangered}{0} & \\ \hline &3&-16&21&\color{orangered}{4}& \end{array} $$

Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 4 } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{0}&3&-16&21&4&-12\\& & 0& 0& 0& \color{blue}{0} \\ \hline &3&-16&21&\color{blue}{4}& \end{array} $$

Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 0 } = \color{orangered}{ -12 } $

$$ \begin{array}{c|rrrrr}0&3&-16&21&4&\color{orangered}{ -12 }\\& & 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{3}&\color{blue}{-16}&\color{blue}{21}&\color{blue}{4}&\color{orangered}{-12} \end{array} $$

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

Synthetic Division Calculator