Polynomial of Degree 2 or 3 over Field Has No Roots If and Only If It is Irreducible

Theorem

A degree \(2\) or \(3\) polynomial \(f \in \mathbb{F}[x]\) for some field \(\mathbb{F}\) has no roots if and only if it is irreducible.

Proof

Suppose \(f \in \mathbb{F}[X]\) factorises non-trivially as \(f = gh\). Given we are working over a field, the the degrees are such that

\[ \deg(f) = \deg(g) + \deg(h).\]

Given our factorisation is non-trivial, we have \(\deg(g), \deg(h) \geq 1\), and hence if \(\deg(f) = \{2, 3\}\), at least one of \(g\) or \(h\) must be degree \(1\).

Given \(f\) has a factor of degree \(1\), it has a root, since \(aX + b = 0\) has general solution \(X = \frac{-b}{a}\).

This correspondence does not generalise to higher degree polynomials.

Example

The polynomial

\[ X^4 + 1 = (X^2 + 2X + 2)(X^2 - 2X + 2)\]

is reducible over \(\mathbb{R}\) but has no roots in \(\mathbb{R}\).