Suppose that $m(t)$ is irreducible over $K$ and $\alpha$ has minimal polynomial $m(t)$ over $K$. Does $m(t)$ necessarily factorise over $K(\alpha)$ into linear (Degree 1) polynomials?
Thinking of concrete examples, I'm trying $K=\mathbb{Q}$ and $\alpha$ is the real cube root of 2...so I want to say no. Is this the right way to think?
Your counterexample is correct as $\mathbb{Q}[\sqrt[3]{2}]$ is a real field but the other roots of $x^3 - 2$ are nonreal.