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