Question 86020
Factor:
{{{x^2+22x+121}}} Notice that the constant term (121) is a perfect square {{{121 = 11^2}}} which suggests the trinomial might, itself, be a perfect square.

Let's try:
{{{x^2+22x+121 = (x+11)(x+11)}}} You can check this using FOIL.
{{{(x+11)(x+11) = x^2+11x+11x+121}}} Simplify.
{{{(x+11)^2 = x^2+22x+121}}}
The factors are:
{{{(x+11)^2}}}