#x^2-3\cdot x-28=# \((x+4)\cdot(x-7)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2-3\cdot x-28# 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:
\[ x^2-(p+q)\cdot x+p\cdot q = x^2-3\cdot x-28\tiny\]
A comparison with #x^2-3\cdot x-28# gives \[
\lineqs{p+q &=& 3\cr p\cdot q &=& -28}\] If #p# and #q# are integers, they are divisors of #-28#. We go through all possible divisors #p# with #p^2\le |-28|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{-28}{p}#:
\[\begin{array}{|r|c|l|}
\hline
p&q&{p+q}\\
\hline
1&-28&-27\\ \hline -1&28&27\\ \hline 2&-14&-12\\ \hline -2&14&12\\ \hline 4&-7&-3\\ \hline -4&7&3 \\
\hline
\end{array}\]
The line of the table with #p=-4# and #q=7# is the only one with sum #3#, hence, this is the answer:
\[x^2-3\cdot x-28=(x+4)\cdot(x-7)\tiny.\]