Because $f$ is <irreducible polynomial>[irreducible], the ideal $(f)$ is maximal in the <polynomial ring> $k[X]$. Hence
$$
F=k[X]/(f)
$$
is a <field extension> of $k$. Let $\alpha=X+(f)\in F$. Evaluation in the quotient gives $f(\alpha)=0$, and the <factor theorem> in $F[X]$ gives
$$
\boxed{f(X)=(X-\alpha)g(X)}
$$
for some $g\in F[X]$. This is the standard <adjoining a root of an irreducible polynomial> construction.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 2 / 11G / a / Solution

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For a polynomial $f$ over a field, $f(a)=0$ exactly when $X-a$ divides $f(X)$. Polynomial division gives $f(X)=(X-a)q(X)+f(a)$.

= Roots of a polynomial
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 Incoming links: Factor theorem