Answer
$c=121$
$(x+11)^{2}$
Work Step by Step
$ x^{2}+22x+c\qquad$ ... find half the coefficient of $x$.
$\displaystyle \frac{22}{2}=11$
$\qquad$ ...square the result.
$ 11^{2}=121\qquad$ ...substitute $c$ with $121$ in the original expression
$x^{2}+22x+121=(x+11)^{2}$
The trinomial $x^{2}+22x+c$ is a perfect square when $c=121.$
Then $x^{2}+22x+121=(x+11)(x+11)=(x+11)^{2}$
