#x^2-10\cdot x+21=# \((x-3)\cdot(x-7)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2-10\cdot x+21# 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:
\[ x^2-(p+q)\cdot x+p\cdot q = x^2-10\cdot x+21\tiny\]
A comparison with #x^2-10\cdot x+21# gives \[
\lineqs{p+q &=& 10\cr p\cdot q &=& 21}\] If #p# and #q# are integers, they are divisors of #21#. We go through all possible divisors #p# with #p^2\le |21|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{21}{p}#:
\[\begin{array}{|r|c|l|}
\hline
p&q&{p+q}\\
\hline
1&21&22\\ \hline -1&-21&-22\\ \hline 3&7&10\\ \hline -3&-7&-10 \\
\hline
\end{array}\]
The line of the table with #p=3# and #q=7# is the only one with sum #10#, hence, this is the answer:
\[x^2-10\cdot x+21=(x-3)\cdot(x-7)\tiny.\]