Check if $ 3 $ is a factor of $ 11x^{2}+16x+6 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}0&11&16&6\\& & 0& \color{black}{0} \\ \hline &\color{blue}{11}&\color{blue}{16}&\color{orangered}{6} \end{array} $$Because the remainder $ \left( \color{red}{ 6 } \right) $ is not zero, we conclude that the $ x $ is not a factor of $ 11x^{2}+16x+6$.
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 at the left.
$$ \begin{array}{c|rrr}\color{blue}{0}&11&16&6\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}0&\color{orangered}{ 11 }&16&6\\& & & \\ \hline &\color{orangered}{11}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 11 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrr}\color{blue}{0}&11&16&6\\& & \color{blue}{0} & \\ \hline &\color{blue}{11}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 0 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrr}0&11&\color{orangered}{ 16 }&6\\& & \color{orangered}{0} & \\ \hline &11&\color{orangered}{16}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 16 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrr}\color{blue}{0}&11&16&6\\& & 0& \color{blue}{0} \\ \hline &11&\color{blue}{16}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 0 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrr}0&11&16&\color{orangered}{ 6 }\\& & 0& \color{orangered}{0} \\ \hline &\color{blue}{11}&\color{blue}{16}&\color{orangered}{6} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 6 }\right)$.