Check if $ x+2 $ is a factor of $ x^{3}+10x^{2}+19x+11 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&1&10&19&11\\& & -2& -16& \color{black}{-6} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{3}&\color{orangered}{5} \end{array} $$Because the remainder $ \left( \color{red}{ 5 } \right) $ is not zero, we conclude that the $ x+2 $ is not a factor of $ x^{3}+10x^{2}+19x+11$.
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 + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&10&19&11\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 1 }&10&19&11\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&10&19&11\\& & \color{blue}{-2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}-2&1&\color{orangered}{ 10 }&19&11\\& & \color{orangered}{-2} & & \\ \hline &1&\color{orangered}{8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 8 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&10&19&11\\& & -2& \color{blue}{-16} & \\ \hline &1&\color{blue}{8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-2&1&10&\color{orangered}{ 19 }&11\\& & -2& \color{orangered}{-16} & \\ \hline &1&8&\color{orangered}{3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&10&19&11\\& & -2& -16& \color{blue}{-6} \\ \hline &1&8&\color{blue}{3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-2&1&10&19&\color{orangered}{ 11 }\\& & -2& -16& \color{orangered}{-6} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{3}&\color{orangered}{5} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 5 }\right)$.