Alphabeta Math
TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-11
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A polynomial of degree two or three over a field is irreducible exactly when it has no root in the field

Statement

Let F be a field and let f∈F[x] have degree 2 or 3. Then f is irreducible over F if and only if f has no root in F.

Facts & Assumptions

Given: A field F and a polynomial f∈F[x] of degree 2 or 3.

[L1]

An element a∈F is a root of f exactly when x−a divides f (Factor theorem over a commutative ring).

[L2]

Degrees add in a product of nonzero polynomials over a field (Over an integral domain, degrees add under multiplication of nonzero polynomials).

[L4]

A nonzero nonunit is irreducible exactly when every factorization has a unit factor (Irreducible and prime elements of an integral domain).

Proof

technique · direct
1.1

If f has a root a, then [L1] gives f=(x−a)q; [L2] and deg⁡f≥2 make both factors nonunits by [L3], so [L4] shows that f is reducible.

givenL1L2L3L4
2.1

Conversely, if f=gh with both factors nonunits, [L2] and [L3] give positive degrees summing to 2 or 3, so one factor has degree 1; writing it as cx+d with c≠0, it has root −c−1d, and that root is a root of f. Thus reducibility implies a root, proving the biconditional.

givenL2L3L4algebra∎

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Dependency tree · two levels

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Sources