No. This is a consequence of the Chebotarev density theorem. To see how it follows, look at exercise 6 at the end of Cassels and Frohlich's "Algebraic Number Theory".

Briefly, the Chebotarev density theorem says that for a Galois extension of global fields $L/K$ and for a finite set $S$ of places of $K$, the proportion of primes of $K$ splitting in $L$ is $1/[L:K]$. If $G=\text{Gal}(L/K)$ and $E$ is the fixed field of some $H\subset G$, it is possible to show that the proportion of places of $K$ with a split factor in $E$ is $|\bigcup_{\rho\in G}\rho H\rho^{-1}|/|G|$, and a lemma on finite groups says that this quotient is not $1$ unless $H=G$.

In your case, take $E=K[x]/(f)$ and $L$ to be a normal extension of $K$ containing $E$. Then "$v$ has a split factor" means "$f$ has a root in the completion $K_v$". If $f$ has a root in each completion (or even a set of completions with density $1$, which includes the case "all but finitely many"), we must have $H=G$ and $E=K$. So $f$ already had a root in $K$.

Edit: I see I missed the note that $K$ might not be finite over $\mathbb{Q}$. This answer is not correct for $K$ of infinite degree over $\mathbb{Q}$, as in FC's answer above.

No. This is a consequence of the Chebotarev density theorem. To see how it follows, look at exercise 6 at the end of Cassels and Frohlich's "Algebraic Number Theory".

Briefly, the Chebotarev density theorem says that for a Galois extension of global fields $L/K$ and for a finite set $S$ of places of $K$, the proportion of primes of $K$ splitting in $L$ is $1/[L:K]$. If $G=\text{Gal}(L/K)$ and $E$ is the fixed field of some $H\subset G$, it is possible to show that the proportion of places of $K$ with a split factor in $E$ is $|\bigcup_{\rho\in G}\rho H\rho^{-1}|/|G|$, and a lemma on finite groups says that this quotient is not $1$ unless $H=G$.

In your case, take $E=K[x]/(f)$ and $L$ to be a normal extension of $K$ containing $E$. Then "$v$ has a split factor" means "$f$ has a root in the completion $K_v$". If $f$ has a root in each completion (or even a set of completions with density $1$, which includes the case "all but finitely many"), we must have $H=G$ and $E=K$. So $f$ already had a root in $K$.

No. This is a consequence of the Chebotarev density theorem. To see how it follows, look at exercise 6 at the end of Cassels and Frohlich's "Algebraic Number Theory".

Briefly, the Chebotarev density theorem says that for a Galois extension of global fields $L/K$ and for a finite set $S$ of places of $K$, the proportion of primes of $K$ splitting in $L$ is $1/[L:K]$. If $G=\text{Gal}(L/K)$ and $E$ is the fixed field of some $H\subset G$, it is possible to show that the proportion of places of $K$ with a split factor in $E$ is $|\cup_{\rho\in G}\rho H\rho^{-1}|/|G|$, and a lemma on finite groups says that this quotient is not 1 unless $H=G$.

In your case, take $E=K[x]/(f)$ and $L$ to be a normal extension of $K$ containing $E$. Then "v has a split factor" means "f has a root in the completion $K_v$". If f has a root in each completion (or even a set of completions with density 1, which includes the case "all but finitely many"), we must have $H=G$ and $E=K$. So f already had a root in $K$.

No. This is a consequence of the Chebotarev density theorem. To see how it follows, look at exercise 6 at the end of Cassels and Frohlich's "Algebraic Number Theory".

Briefly, the Chebotarev density theorem says that for a Galois extension of global fields $L/K$ and for a finite set $S$ of places of $K$, the proportion of primes of $K$ splitting in $L$ is $1/[L:K]$. If $G=\text{Gal}(L/K)$ and $E$ is the fixed field of some $H\subset G$, it is possible to show that the proportion of places of $K$ with a split factor in $E$ is $|\bigcup_{\rho\in G}\rho H\rho^{-1}|/|G|$, and a lemma on finite groups says that this quotient is not $1$ unless $H=G$.

In your case, take $E=K[x]/(f)$ and $L$ to be a normal extension of $K$ containing $E$. Then "$v$ has a split factor" means "$f$ has a root in the completion $K_v$". If $f$ has a root in each completion (or even a set of completions with density $1$, which includes the case "all but finitely many"), we must have $H=G$ and $E=K$. So $f$ already had a root in $K$.