Factoring $x^2+4$ [Elementary]

Question

The real factors of $ x^2+4$ are:

How to factor this without being complex?

Answer 1

Hint: we have $$x^2+4=x^2\color{blue}{+4x}+4\color{blue}{-4x}=(x+2)^2-(2\sqrt{x})^2$$ which can be factored as a difference of squares.

The answer would be

$$x^2+4=(x-2\sqrt{x}+2)(x+2\sqrt{x}+2)$$

Answer 2

The factorization of a polynomial in monic binomials is unique. As $x^2+4=(x+2i)(x-2i)$, there is no other option.