In general, [tex]a^2- b^2= (a- b)(a+ b)/[tex]
Actually, $\displaystyle 2x^2- 121= (\sqrt{2}x)^2- 11^2= (\sqrt{2}x- 11)(\sqrt{2}x+ 11)$, not $\displaystyle (x- 11)(x+ 11)$. I presume that was a mis-type.
$\displaystyle 4x^2= (2x)^2$ so, yes, $\displaystyle 4x^2- 121= (2x- 11)(2x+ 11)$.