#x^2-5\cdot x-36=# \((x+4)\cdot(x-9)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2-5\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:
\[ x^2-(p+q)\cdot x+p\cdot q = x^2-5\cdot x-36\tiny\]
A comparison with #x^2-5\cdot x-36# gives \[
\lineqs{p+q &=& 5\cr p\cdot q &=& -36}\] 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}#:
\[\begin{array}{|r|c|l|}
\hline
p&q&{p+q}\\
\hline
1&-36&-35\\ \hline -1&36&35\\ \hline 2&-18&-16\\ \hline -2&18&16\\ \hline 3&-12&-9\\ \hline -3&12&9\\ \hline 4&-9&-5\\ \hline -4&9&5\\ \hline 6&-6&0\\ \hline -6&6&0 \\
\hline
\end{array}\]
The line of the table with #p=-4# and #q=9# is the only one with sum #5#, hence, this is the answer:
\[x^2-5\cdot x-36=(x+4)\cdot(x-9)\tiny.\]