#x^2+4\cdot x+3=# \((x+1)\cdot(x+3)\)

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