Answer
$\displaystyle \frac{x}{x+2},\qquad x\neq-2$
Work Step by Step
We factor what we can.
Factoring $ x^{2}+bx+c $, we search for factors (m and n) of $ c $ whose sum is $ b.$
If they (m and n) exist, we obtain $(x+m)(x+n)$.
Denominator $ = (x+2)(x+2)$, so $+2$ and $+2$ are factors of $4$, and their sum is $+4.$
(we could have also recognized a square of a sum).
Numerator = $ x(x+2)$
The greatest common factor is x, which we factor out.
Problem expression = $\displaystyle \frac{x(x+2)}{ (x+2)(x+2)},\qquad x\neq-2$
All numbers producing zero in the denominator are excluded from the domain.
The common factors cancel, so the expression reduces to:
= $\displaystyle \frac{x}{x+2},\qquad x\neq-2$
