#x^2+13\cdot x+36=# \((x+4)\cdot(x+9)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2+13\cdot x+36# can be written as #(x-p)\cdot(x-q)#. If the absolute value of #p# is greater than #q#, we interchange them, so #|p|\le |q|#. We expand the brackets and compare the result with the original expression:
\[ x^2-(p+q)\cdot x+p\cdot q = x^2+13\cdot x+36\tiny\]
A comparison with #x^2+13\cdot x+36# gives \[
\lineqs{p+q &=& -13\cr p\cdot q &=& 36}\] If #p# and #q# are integers, they are divisors of #36#. We go through all possible divisors #p# with #p^2\le |36|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{36}{p}#:
\[\begin{array}{|r|c|l|}
\hline
p&q&{p+q}\\
\hline
1&36&37\\ \hline -1&-36&-37\\ \hline 2&18&20\\ \hline -2&-18&-20\\ \hline 3&12&15\\ \hline -3&-12&-15\\ \hline 4&9&13\\ \hline -4&-9&-13\\ \hline 6&6&12\\ \hline -6&-6&-12 \\
\hline
\end{array}\]
The line of the table with #p=-4# and #q=-9# is the only one with sum #-13#, hence, this is the answer:
\[x^2+13\cdot x+36=(x+4)\cdot(x+9)\tiny.\]