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