#x^2-x-72=# \((x+8)\cdot(x-9)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2-x-72# 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:
\[ x^2-(p+q)\cdot x+p\cdot q = x^2-x-72\tiny\]
A comparison with #x^2-x-72# gives \[
\lineqs{p+q &=& 1\cr p\cdot q &=& -72}\] If #p# and #q# are integers, they are divisors of #-72#. We go through all possible divisors #p# with #p^2\le |-72|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{-72}{p}#:
\[\begin{array}{|r|c|l|}
\hline
p&q&{p+q}\\
\hline
1&-72&-71\\ \hline -1&72&71\\ \hline 2&-36&-34\\ \hline -2&36&34\\ \hline 3&-24&-21\\ \hline -3&24&21\\ \hline 4&-18&-14\\ \hline -4&18&14\\ \hline 6&-12&-6\\ \hline -6&12&6\\ \hline 8&-9&-1\\ \hline -8&9&1 \\
\hline
\end{array}\]
The line of the table with #p=-8# and #q=9# is the only one with sum #1#, hence, this is the answer:
\[x^2-x-72=(x+8)\cdot(x-9)\tiny.\]