Factorization over $\mathbb{R}$

Over $\mathbb{R}$.
$$X^4-1=(X^2-1)(X^2+1)=(X-1)(X+1)(X^2+1).$$
$X^2+1$ is irreducible over $\mathbb{R}$.

Over $\mathbb{C}$. $X^2+1=(X-i)(X+i)$, so
$$X^4-1=(X-1)(X+1)(X-i)(X+i).$$

Conclusion. Over $\mathbb{R}$: $(X-1)(X+1)(X^2+1)$. Over $\mathbb{C}$: four linear factors.