Irreducible polynomials and primitive elements over a finite field

I am working on generating cyclic codes using primitive elements of finite fields.

I am seeking to answer this question: Is it true that for an irreducible polynomial $f(x)$ over a finite field $F$ of order $q$, in the splitting field $K$ (of order $q^t$) over $F$, either every root of $f(x)$ is a primitive element of $K$, or all roots of $f(x)$ are not primitive elements?

My thinking thus far is that cyclotomic cosets could be used. It is not hard to show that $\gcd(kq^r,q^t-1) = 1$ as long as $\gcd(k,q^t-1)=1$. Hence, if $\alpha$ is a primitive root of $K$, then $\alpha^{kq^r}$ is primitive as long as $\alpha^k$ is primitive. I'm not sure I'm on the right track here.

Answer

Yes, if $f$ is irreducible, then all its zeroes are in the same orbit
of the action of the Galois group of $K/F$. As the Galois group consists of
field automorphisms, then if one zero is a primitive element, then they all
are.