The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&1&6&11&6\\& & -1& -5& \color{black}{-6} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{6}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x+1 $ is a factor of the $ x^{3}+6x^{2}+11x+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 from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&6&11&6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 1 }&6&11&6\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 1 } = \color{blue}{ -1 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&6&11&6\\& & \color{blue}{-1} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-1&1&\color{orangered}{ 6 }&11&6\\& & \color{orangered}{-1} & & \\ \hline &1&\color{orangered}{5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 5 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&6&11&6\\& & -1& \color{blue}{-5} & \\ \hline &1&\color{blue}{5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}-1&1&6&\color{orangered}{ 11 }&6\\& & -1& \color{orangered}{-5} & \\ \hline &1&5&\color{orangered}{6}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 6 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&6&11&6\\& & -1& -5& \color{blue}{-6} \\ \hline &1&5&\color{blue}{6}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-1&1&6&11&\color{orangered}{ 6 }\\& & -1& -5& \color{orangered}{-6} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{6}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.