#x^2-2\cdot x-63=# \((x+7)\cdot(x-9)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2-2\cdot x-63# 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:
\[ x^2-(p+q)\cdot x+p\cdot q = x^2-2\cdot x-63\tiny\]
A comparison with #x^2-2\cdot x-63# gives \[
\lineqs{p+q &=& 2\cr p\cdot q &=& -63}\] If #p# and #q# are integers, they are divisors of #-63#. We go through all possible divisors #p# with #p^2\le |-63|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{-63}{p}#:
\[\begin{array}{|r|c|l|}
\hline
p&q&{p+q}\\
\hline
1&-63&-62\\ \hline -1&63&62\\ \hline 3&-21&-18\\ \hline -3&21&18\\ \hline 7&-9&-2\\ \hline -7&9&2 \\
\hline
\end{array}\]
The line of the table with #p=-7# and #q=9# is the only one with sum #2#, hence, this is the answer:
\[x^2-2\cdot x-63=(x+7)\cdot(x-9)\tiny.\]