Check if $ x-5 $ is a factor of $ x^{2}+2x-36 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}5&1&2&-36\\& & 5& \color{black}{35} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{orangered}{-1} \end{array} $$Because the remainder $ \left( \color{red}{ -1 } \right) $ is not zero, we conclude that the $ x-5 $ is not a factor of $ x^{2}+2x-36$.
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 -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrr}\color{blue}{5}&1&2&-36\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}5&\color{orangered}{ 1 }&2&-36\\& & & \\ \hline &\color{orangered}{1}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrr}\color{blue}{5}&1&2&-36\\& & \color{blue}{5} & \\ \hline &\color{blue}{1}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 5 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrr}5&1&\color{orangered}{ 2 }&-36\\& & \color{orangered}{5} & \\ \hline &1&\color{orangered}{7}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 7 } = \color{blue}{ 35 } $.
$$ \begin{array}{c|rrr}\color{blue}{5}&1&2&-36\\& & 5& \color{blue}{35} \\ \hline &1&\color{blue}{7}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -36 } + \color{orangered}{ 35 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrr}5&1&2&\color{orangered}{ -36 }\\& & 5& \color{orangered}{35} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{orangered}{-1} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -1 }\right)$.