= Solution
Take $\boxed{K=F[x]/(f)}$ and let $\alpha$ be the <residue> class of $x$. Division with remainder represents every class by a <polynomial> of degree less than $n$, so every element is a <polynomial> in $\alpha$ and $f(\alpha)=0$. A nonzero class represented by $u$ has $\gcd(u,f)=1$ because $f$ is irreducible. Bézout's identity gives $au+bf=1$, making the class of $a$ its inverse. Thus $K$ is a <field>, containing $F$ through its constant classes.
A <splitting field> of a nonconstant <polynomial> is an extension in which it factors into linear factors and which is generated over the original <field> by its roots. To construct one, choose an irreducible factor and use the preceding quotient construction to adjoin a root. Divide off a linear factor in that extension and repeat on the remaining <polynomial>. Each stage lowers the remaining degree, so finitely many stages suffice. The <field> generated by all roots in the resulting <field> is the required <splitting field>. This induction works with repeated roots and does not assume separability. The two roman clauses in the source specify properties of this one construction; they are not separate questions.
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