Check if $ 3 $ is a factor of $ 2x^{3}+11x^{2}+16x+6 $.

The synthetic division table is:

$$ \begin{array}{c|rrrr}0&2&11&16&6\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{2}&\color{blue}{11}&\color{blue}{16}&\color{orangered}{6} \end{array} $$

Because the remainder $ \left( \color{red}{ 6 } \right) $ is not zero, we conclude that the $ x $ is not a factor of $ 2x^{3}+11x^{2}+16x+6$.

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.

$$ \begin{array}{c|rrrr}\color{blue}{0}&2&11&16&6\\& & & & \\ \hline &&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}0&\color{orangered}{ 2 }&11&16&6\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}\color{blue}{0}&2&11&16&6\\& & \color{blue}{0} & & \\ \hline &\color{blue}{2}&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ 0 } = \color{orangered}{ 11 } $

$$ \begin{array}{c|rrrr}0&2&\color{orangered}{ 11 }&16&6\\& & \color{orangered}{0} & & \\ \hline &2&\color{orangered}{11}&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 11 } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrr}\color{blue}{0}&2&11&16&6\\& & 0& \color{blue}{0} & \\ \hline &2&\color{blue}{11}&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 0 } = \color{orangered}{ 16 } $

$$ \begin{array}{c|rrrr}0&2&11&\color{orangered}{ 16 }&6\\& & 0& \color{orangered}{0} & \\ \hline &2&11&\color{orangered}{16}& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 16 } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrr}\color{blue}{0}&2&11&16&6\\& & 0& 0& \color{blue}{0} \\ \hline &2&11&\color{blue}{16}& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 0 } = \color{orangered}{ 6 } $

$$ \begin{array}{c|rrrr}0&2&11&16&\color{orangered}{ 6 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{2}&\color{blue}{11}&\color{blue}{16}&\color{orangered}{6} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ 6 }\right)$.

Synthetic Division Calculator