#x^2+14\cdot x+48=# \((x+6)\cdot(x+8)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2+14\cdot x+48# 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:
\[ x^2-(p+q)\cdot x+p\cdot q = x^2+14\cdot x+48\tiny\]
A comparison with #x^2+14\cdot x+48# gives \[
\lineqs{p+q &=& -14\cr p\cdot q &=& 48}\] If #p# and #q# are integers, they are divisors of #48#. We go through all possible divisors #p# with #p^2\le |48|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{48}{p}#:
\[\begin{array}{|r|c|l|}
\hline
p&q&{p+q}\\
\hline
1&48&49\\ \hline -1&-48&-49\\ \hline 2&24&26\\ \hline -2&-24&-26\\ \hline 3&16&19\\ \hline -3&-16&-19\\ \hline 4&12&16\\ \hline -4&-12&-16\\ \hline 6&8&14\\ \hline -6&-8&-14 \\
\hline
\end{array}\]
The line of the table with #p=-6# and #q=-8# is the only one with sum #-14#, hence, this is the answer:
\[x^2+14\cdot x+48=(x+6)\cdot(x+8)\tiny.\]