Is $x^3 + x^2 + x + 1$ irreducible over set of reals?

Question

$x^3 + x^2 + x + 1$ = ($x^2$ + 1)($x$ + 1)

I know that both are non-units.

So, since it splits into factors, then it is reducible over reals.

Is my theory correct?

Answer 1

In the interests of having an answer instead of only comments:

@Surb is right. For $p$ to be irreducible over $\Bbb R$ means no $q\in\Bbb R[X]$ with $1\le\deg q<\deg p$ satisfies $q|p$; it's nothing to do with which roots of $p$ are elements of $\Bbb R$. See also @PacoAdajar's comparison of reducibility to a different question, that of your polynomial's splitting field, which is $\Bbb Q[i]$.