Check if $ x+3 $ is a factor of $ 2x^{3}+11x^{2}+16x+6 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&2&11&16&6\\& & -6& -15& \color{black}{-3} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{1}&\color{orangered}{3} \end{array} $$Because the remainder $ \left( \color{red}{ 3 } \right) $ is not zero, we conclude that the $ x+3 $ is not a factor of $ 2x^{3}+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 from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&11&16&6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 2 }&11&16&6\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&11&16&6\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-3&2&\color{orangered}{ 11 }&16&6\\& & \color{orangered}{-6} & & \\ \hline &2&\color{orangered}{5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&11&16&6\\& & -6& \color{blue}{-15} & \\ \hline &2&\color{blue}{5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}-3&2&11&\color{orangered}{ 16 }&6\\& & -6& \color{orangered}{-15} & \\ \hline &2&5&\color{orangered}{1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&11&16&6\\& & -6& -15& \color{blue}{-3} \\ \hline &2&5&\color{blue}{1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-3&2&11&16&\color{orangered}{ 6 }\\& & -6& -15& \color{orangered}{-3} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{1}&\color{orangered}{3} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 3 }\right)$.