= Solution
Proceed by induction on $d=\deg f$. A nonconstant <polynomial> over a <field> has an irreducible factor $h$. Part (a) gives an extension $k_1/k$ containing a root $\alpha_1$ of $h$, hence of $f$. The factor theorem gives
$$
f(X)=(X-\alpha_1)f_1(X),
\qquad f_1\in k_1[X],
$$
where $f_1$ is monic of degree $d-1$. Apply the induction hypothesis over $k_1$ and compose the resulting field extensions. The final field $F$ contains roots $\alpha_1,\ldots,\alpha_d$ and
$$
\boxed{f(X)=\prod_{i=1}^d(X-\alpha_i)}.
$$
Equivalently, every polynomial has a <splitting field>.
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