Let $E$ be a real vector space of finite dimension $n$ and $f \neq 0$ an endomorphism of $E$. I proved that there exist a real polynomial $P$ such that $P(f) = 0$. Now, we assume that $P$ has no real roots. Let $Q$ be an irreducible real polynomial $Q$ of degree $2$ which devides $P$. How can I prove that $\mathrm{ker}(Q(f)) \neq \left\{ 0 \right\}$ without using the characteristic polynomial of $f$ ?
I can write $Q(X) = (X-\mu)(X-\overline{\mu})$ where $\mu \in \mathbb{C}$. Because $\det Q(f) = \det(f-\mu \, \mathrm{Id}) \det( f - \overline{\mu} \, \mathrm{Id})$, I need to prove that $\det(f-\mu \, \mathrm{Id}) = 0$ but I don't see how to prove this without using the characteristic polynomial of $f$.
You know $P(f)=0$. Now write $P$ as a product of irreducible $Q_i$, $i=1\dots r$, $$ P = Q_1 \cdot \dots \cdot Q_r. $$ Then you have $$ 0= P(f) = Q_1(f) \circ \dots \circ Q_r(f). $$ This implies that one of the factors $Q_i$ is not injective (otherwise the mapping $Q_1(f) \circ \dots \circ Q_r(f)$ would be injective, a contradiction to being zero). This gives you $\ker Q_j(f)\ne \{0\}$.
Note that you cannot prove that one specific factor $Q_j$ yields $\ker Q_j(f)\ne \{0\}$. The property $P(f)=0$ only gives you that one of these factors is not injective.