Let \(k\) be a field and let \(f(x) \in k[x]\) be a polynomial of degree \(\deg(f) \in \{2, 3\}\). Prove that \(f(x)\) is irreducible in \(k[x]\) if and only if \(f(x)\) has no roots in \(k\).
Solution
If \(f\) is irreducible and had a root \(a\in k\), then the factor theorem would give \[ f(x)=(x-a)g(x) \] with \(\deg g=\deg f-1\ge1\), a nontrivial factorization. Hence an irreducible polynomial of degree \(2\) or \(3\) has no root in \(k\).
Conversely, suppose \(f\) is reducible. Then \[ f=gh \] with \(\deg g,\deg h\ge1\) and \[ \deg g+\deg h=\deg f\in\{2,3\}. \] Therefore at least one factor has degree \(1\). Write that factor as \(ax+b\) with \(a\ne0\). It vanishes at \(-b/a\in k\), so \(f\) has a root in \(k\).
Thus, for degrees \(2\) and \(3\), \[ f\text{ irreducible}\quad\Longleftrightarrow\quad f\text{ has no root in }k. \]