#x^2-6\cdot x-16=# \((x+2)\cdot(x-8)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2-6\cdot x-16# 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:
\[ x^2-(p+q)\cdot x+p\cdot q = x^2-6\cdot x-16\tiny\]
A comparison with #x^2-6\cdot x-16# gives \[
\lineqs{p+q &=& 6\cr p\cdot q &=& -16}\] If #p# and #q# are integers, they are divisors of #-16#. We go through all possible divisors #p# with #p^2\le |-16|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{-16}{p}#:
\[\begin{array}{|r|c|l|}
\hline
p&q&{p+q}\\
\hline
1&-16&-15\\ \hline -1&16&15\\ \hline 2&-8&-6\\ \hline -2&8&6\\ \hline 4&-4&0\\ \hline -4&4&0 \\
\hline
\end{array}\]
The line of the table with #p=-2# and #q=8# is the only one with sum #6#, hence, this is the answer:
\[x^2-6\cdot x-16=(x+2)\cdot(x-8)\tiny.\]