#x^2+15\cdot x+54=# \((x+6)\cdot(x+9)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2+15\cdot x+54# 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:
\[ x^2-(p+q)\cdot x+p\cdot q = x^2+15\cdot x+54\tiny\]
A comparison with #x^2+15\cdot x+54# gives \[
\lineqs{p+q &=& -15\cr p\cdot q &=& 54}\] If #p# and #q# are integers, they are divisors of #54#. We go through all possible divisors #p# with #p^2\le |54|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{54}{p}#:
\[\begin{array}{|r|c|l|}
\hline
p&q&{p+q}\\
\hline
1&54&55\\ \hline -1&-54&-55\\ \hline 2&27&29\\ \hline -2&-27&-29\\ \hline 3&18&21\\ \hline -3&-18&-21\\ \hline 6&9&15\\ \hline -6&-9&-15 \\
\hline
\end{array}\]
The line of the table with #p=-6# and #q=-9# is the only one with sum #-15#, hence, this is the answer:
\[x^2+15\cdot x+54=(x+6)\cdot(x+9)\tiny.\]