#x^2-12\cdot x+32=# \((x-4)\cdot(x-8)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2-12\cdot x+32# 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:
\[ x^2-(p+q)\cdot x+p\cdot q = x^2-12\cdot x+32\tiny\]
A comparison with #x^2-12\cdot x+32# gives \[
\lineqs{p+q &=& 12\cr p\cdot q &=& 32}\] If #p# and #q# are integers, they are divisors of #32#. We go through all possible divisors #p# with #p^2\le |32|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{32}{p}#:
\[\begin{array}{|r|c|l|}
\hline
p&q&{p+q}\\
\hline
1&32&33\\ \hline -1&-32&-33\\ \hline 2&16&18\\ \hline -2&-16&-18\\ \hline 4&8&12\\ \hline -4&-8&-12 \\
\hline
\end{array}\]
The line of the table with #p=4# and #q=8# is the only one with sum #12#, hence, this is the answer:
\[x^2-12\cdot x+32=(x-4)\cdot(x-8)\tiny.\]