Is there a field extension that causes polynomials to factor, but not into linear factors?

Question

Does there exist a field extension $E/F$ such that:

1. There is a irreducible polynomial $P \in F[x]$ that becomes reducible when viewed as an element of $E[x]$.
2. Every $P \in F[x]$ with no roots in $F$ also has no roots in $E$.

Answer 1

Every element $a \in E\backslash F$ would have to be transcendental, since if $a$ was algebraic over $F$ and a root of a minimal polynomial $f(x)\in F[x]$ then $f$ has no roots in $F[x]$ but has a root $a$ in $E[x]$.

Suppose $f(x) \in F[x]$ is irreducible but $f(x)=P(x)Q(x)$ where $P,Q\in E[x]$. Let $\{a_i\}$ be the set of coefficients of $P$ and $Q$. Then there exists $i$ such that $a_i \in E\backslash F$ so $a_i$ is not algebraic over $F$.

Let $G$ be the splitting field of $f(x)$. Then $P(x),Q(x) \in G[x]$. $G$ is an algebraic extension of $F$ and so all of the coefficients of $P$ and $Q$ are algebraic over $F$. This is a contradiction.