Question

Find the factors of $x^2+4x+1$.

Answer 1

Here we do not see any difference of two squares. Now $x^2+4x$ can be completed to a square by adding $4$ to it. Thus
$x^2+4x+1=x^2+2\left(2\right)\left(x\right)+2^2-2^2={(x+2)}^2-4$
(we added $2^2=4$ and subtracted the same quantity so that nothing is changed) Hence
$x^2+4x+1={(x+2)}^2-4+1={(x+2)}^2-3={(x+2)}^2-{\left(\sqrt{3}\right)}^2$
The important point to be noted here is that $3$ is the square of another real number $\sqrt{3}$. Now we are in a familiar situation. We obtain
$x^2+4x+1={(x+2)}^2-{\left(\sqrt{3}\right)}^2=(x+2+\sqrt{3})(x+2-\sqrt{3})$