#x^2-x-42=# \((x+6)\cdot(x-7)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2-x-42# 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:
\[ x^2-(p+q)\cdot x+p\cdot q = x^2-x-42\tiny\]
A comparison with #x^2-x-42# gives \[
\lineqs{p+q &=& 1\cr p\cdot q &=& -42}\] If #p# and #q# are integers, they are divisors of #-42#. We go through all possible divisors #p# with #p^2\le |-42|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{-42}{p}#:
\[\begin{array}{|r|c|l|}
\hline
p&q&{p+q}\\
\hline
1&-42&-41\\ \hline -1&42&41\\ \hline 2&-21&-19\\ \hline -2&21&19\\ \hline 3&-14&-11\\ \hline -3&14&11\\ \hline 6&-7&-1\\ \hline -6&7&1 \\
\hline
\end{array}\]
The line of the table with #p=-6# and #q=7# is the only one with sum #1#, hence, this is the answer:
\[x^2-x-42=(x+6)\cdot(x-7)\tiny.\]