In Bourbaki, Algebra, V, 37, we have Proposition 3, which states that: Let $f$ be a non-zero polynomial in $K[X]$ and let $\Omega$ be an algebraically closed extension of $K$. The following conditions are equivalent:
c) There exists an extension $L$ of $K$ such that $f$ splits in $L[X]$ into a product of distinct polynomials of degree $\leq 1$.
d) The roots of $f$ in $\Omega$ are simple.
I am very confused by c). Let us consider $f(X) = (2X-2)(X-1) \in \mathbb{R}[X]$: We have that $f$ satisfies c) but not d). The proof by Bourbaki to show that c) implies d) begins as follows: "Under the hypothesis c) there exist an element $\lambda \not= 0$ in $L$ and distinct elements $\alpha_1, \ldots, \alpha_n$ of $L$ such that $f(X) = \lambda (X - \alpha_1) ... (X - \alpha_n)$." With my $f$, we have $f(X) = 2 (X-1) (X-1)$ and the $\alpha_i$'s are not distinct. Is it a mistake in Bourbaki ("monic" is missing) or did I misunderstand something here?