Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A real polynomial vanishing at aa is divisible by x−ax-a

Statement

If p(x)=∑k<nakxkp(x)=\sum_{k<n}a_kx^k and p(a)=0p(a)=0, then there is a real polynomial qq with p(x)=(x−a)q(x)p(x)=(x-a)q(x) for every real xx. The conventions and prerequisite facts used below are recorded in Formal real polynomials, evaluation, degree, leading coefficient, and monic polynomials, Factorisation of bn−anb^n - a^n, and the resulting Lipschitz estimate, Laws of finite sums and finite products.

Facts & Assumptions

Given: A polynomial pp and a real root aa.

Proof

technique · constructive
1.1

For each k≥1k\ge1, the power-difference factorization gives xk−ak=(x−a)∑j<kajxk−1−jx^k-a^k=(x-a)\sum_{j<k}a^jx^{k-1-j}.

given
1.2

Since p(a)=0p(a)=0, write p(x)=∑k<nak(xk−ak)p(x)=\sum_{k<n}a_k(x^k-a^k).

algebra
2.1

Substitute the factorization from step 1.1 and collect the finite coefficient sums into a polynomial qq.

constructdischarge-construct∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 37 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources