#x^2-16\cdot x+63=# \((x-7)\cdot(x-9)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2-16\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:
\[ x^2-(p+q)\cdot x+p\cdot q = x^2-16\cdot x+63\tiny\]
A comparison with #x^2-16\cdot x+63# gives \[
\lineqs{p+q &=& 16\cr p\cdot q &=& 63}\] 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}#:
\[\begin{array}{|r|c|l|}
\hline
p&q&{p+q}\\
\hline
1&63&64\\ \hline -1&-63&-64\\ \hline 3&21&24\\ \hline -3&-21&-24\\ \hline 7&9&16\\ \hline -7&-9&-16 \\
\hline
\end{array}\]
The line of the table with #p=7# and #q=9# is the only one with sum #16#, hence, this is the answer:
\[x^2-16\cdot x+63=(x-7)\cdot(x-9)\tiny.\]