#x^2+17\cdot x+72=# \((x+8)\cdot(x+9)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2+17\cdot 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:
\[ x^2-(p+q)\cdot x+p\cdot q = x^2+17\cdot x+72\tiny\]
A comparison with #x^2+17\cdot x+72# gives \[
\lineqs{p+q &=& -17\cr p\cdot q &=& 72}\] 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}#:
\[\begin{array}{|r|c|l|}
\hline
p&q&{p+q}\\
\hline
1&72&73\\ \hline -1&-72&-73\\ \hline 2&36&38\\ \hline -2&-36&-38\\ \hline 3&24&27\\ \hline -3&-24&-27\\ \hline 4&18&22\\ \hline -4&-18&-22\\ \hline 6&12&18\\ \hline -6&-12&-18\\ \hline 8&9&17\\ \hline -8&-9&-17 \\
\hline
\end{array}\]
The line of the table with #p=-8# and #q=-9# is the only one with sum #-17#, hence, this is the answer:
\[x^2+17\cdot x+72=(x+8)\cdot(x+9)\tiny.\]