Is $x^2-2$ irreducible over R and Q?

Question

I'm not sure if it is valid to say that $x^2 - 2$ can be factorised to $2\cdot\left(\frac 12x^2 - 1\right)$ for it to be reducible in Q.

Though I know $(x + \sqrt{2})(x - \sqrt{2})$ works in the reals.

Answer 1

The polynomial is ofcourse reducible over $\mathbb{R}$ as it can be written as a product of polynomials of (strictly) smaller degrees.

Again, by Eisenstein's criteria it is irreducible over $\mathbb{Q}$

Answer 2

Suppose you can factor it as $x^2 - 2 = (x+a)(x+b)$ , $a,b \in \mathbb{Z}$. Then:

$a+b= 0, ab = -2$, thus: $a = - b, -a^2 = -2$.

So: $a^2 = 2$, and this can't happen for $a \in \mathbb{Z}$.