Deep Dive: Factor and Remainder Theorems

Provenance (2026-09-29). Read in full for this pass: research/veridian/WMA12-verified-facts.md (§1–§2 for the unit/spec identity and spec 2.1's verbatim wording; §3 for the formula-booklet absence; §4 for mark-scheme conventions and the M/A/B/dM1/isw/cso/ag glossary; §6.2 for the Jan 2024 Q1 and Oct 2023 Q4 anchors; §7's "verified across at least 4 series" claim; §8's flagged issues; and §9's full lesson-audit log — both independent factor-and-remainder-theorems.ts audit entries dated 2026-09-14, plus the two stationary-points-and-curve-sketching.ts audit entries, for the cross-topic Jan 2021 Q2 anchor); the live lesson file content/veridian/curriculum/wma12/factor-and-remainder-theorems.ts (read in full, all 926 lines — every teach/mechanism block, the worked-chain, chain-drill, all three method-comparison routes, both marked-solution routes including the wrong path, all eight trap-taxonomy items, all six MCQs and the reference card — so nothing below restates its worked content stage-by-stage); content/veridian/technique-hub/wma12/codex-formula-and-method-bank.md (§2's identity table and method-skeleton entry, read in full); content/veridian/technique-hub/wma12/forge-exam-technique- guide.md (read in full — the booklet note, the June 2025 per-question topic/mark table, and the trap cross-reference list); content/veridian/technique-hub/wma12/signal-frequency-and-boundaries.md (read in full — the Part 1 frequency row for this topic); and content/veridian/question-bank/ wma12.ts (grepped for "factor", "remainder", "divi" — no seeded question on this topic exists there, so no overlap to avoid). No new primary-source PDF was fetched for this document; every real-exam fact below is drawn from citations already verified in the sources just listed. Every arithmetic step shown — every substitution, division, and factorisation, including the ones the sources above only state the result of — was independently recomputed by hand (synthetic division checked against an expansion-back check) before being written in, not copied blind. Not affiliated with or endorsed by Pearson Edexcel; every quotation below is short (under ~15 words) and attributed to a named document — no full question, mark-scheme extract or examiner-report passage is reproduced at length.

#CODEX Deep-Dive — Factor and Remainder Theorems (WMA12, spec 2.1)

#What this document is, and isn't

The lesson file already teaches spec 2.1 pedagogically, in full: a mechanism block deriving both theorems from the one division identity; a teach block distinguishing the "remainder given" and "factor given" question shapes; a worked-chain and chain-drill on dividing out a known factor and factorising the quotient; a three-route method-comparison (remainder theorem, long division, compare coefficients) on the same 3-mark question shape; a 9-mark marked-solution combining a "show that" factor condition with a "show that" remainder condition into a simultaneous-equations-then-factorise problem, plus its own worked wrong path; eight trap-taxonomy items, each with a real citation; and six MCQs. codex-formula-and-method-bank.md already holds the compact identity table and a method-skeleton write-up pointing back at the lesson. forge-exam-technique-guide.md already covers the exam-craft side (the booklet-absence note, the June 2025 per-question topic map). None of that is repeated below.

What none of those four documents does is complete the two worked examples the spec's own guidance names for this exact item — "Only division by (ax + b) or (ax − b) will be required" and students "may be required to factorise cubic expressions such as" two specific cubics, quoted verbatim in the lesson's own header comment and in the codex bank's §2, but never actually solved anywhere in this course's material. Nor does any existing document state, as one place, the general "no factor given — search from scratch" mechanism carried all the way through to a genuinely distinct sub-case the spec's own examples happen to contain: a repeated linear factor. Nor does any existing document assemble a single decision framework across the three real methods, or a single consolidated ledger of every real series this specific skill has been confirmed tested in — including two cross-topic sightings that currently sit inside two other lessons' own audit trails and have never been pulled together under this topic before.

The honest evidentiary picture. This is a well-evidenced topic by this course's own standard — the facts bank counts it "verified across at least 4 series" — richer than several other WMA12 sub-topics reviewed elsewhere in this project. This document goes correspondingly deep. But two disciplines are kept throughout: every claim about what has actually been tested names its series and question; and where a genuine mathematical sub-case is included because the theory or the spec's own wording requires it, with no real per-question anchor found for that specific shape, this document says so plainly rather than dressing it in an invented citation. Two labels are used consistently below: [real] for a verified fact from a genuine Pearson document, series and question given, and [original] for a fresh micro-example built to demonstrate a mechanism, with no claim to resemble any specific real Pearson question — the same disclosure convention the lesson file itself uses throughout.

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#Part 1 — The identity, recapped just far enough to stand alone

The lesson's mechanism block derives this in full; it is recapped here only as a foundation for what follows. Dividing $f(x)$ by a linear expression $(ax-b)$ produces some quotient $q(x)$ and some remainder $r$, and that division is exactly the statement of an identity — true for every value of $x$, not a special few — $f(x) \equiv (ax-b)\,q(x) + r$. Because the divisor has degree 1, $r$ must be a constant: anything of degree 1 or more left over could still be divided by $(ax-b)$ again, so the division would not have finished. Substituting the one value that makes the divisor zero, $x = \frac{b}{a}$, deletes the entire quotient term — whatever it is, it is multiplied by zero — and leaves $f\!\left(\frac{b}{a}\right) = r$. That is the remainder theorem. Setting $r = 0$ is not a second fact but the same identity read at its own special case: $(ax-b)$ is a factor of $f(x)$ exactly when $f\!\left(\frac{b}{a}\right) = 0$ — the factor theorem, an "if and only if" because both directions of the argument run down the identical line.

This is also the reason spec 2.1's own restriction exists in the form it does. The general algebraic division of one polynomial by another always satisfies $\deg(\text{remainder}) < \deg(\text{divisor})$ — that much is true whatever the divisor's degree. Restricting WMA12 to a linear divisor, $(ax+b)$ or $(ax-b)$ only, is exactly what collapses "the remainder has degree less than 1" down to "the remainder is a single number" — the entire reason one substitution can pin it down completely, with nothing left to solve for. A quadratic or higher-degree divisor would leave a linear-or-higher remainder, which is a different (and harder) problem this spec item is explicitly built to avoid: "Only division by (ax + b) or (ax − b) will be required" [real, spec P2.3, item 2.1's own guidance, quoted directly in WMA12-verified-facts.md §2].

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#Part 2 — Sub-case: no factor given at all — searching from scratch, completed

The lesson's own mechanism block derives why the search is bounded — if $f(x) \equiv (ax-b)q(x)$ with integer coefficients throughout, then $b$ must divide $f$'s constant term and $a$ must divide $f$'s leading coefficient, because each is the product of the corresponding two coefficients of the divisor and the quotient. What it does not do is carry either of the spec's own two named cubics all the way to a finished factorisation. This section does both, plus a genuine sub-case — a repeated linear factor — that neither cubic alone would have surfaced on its own.

#2a — A real anchor for the repeated-factor sub-case, newly identified here

Before either worked example: the facts bank's own audit log already cites a real mark-scheme quotient for this exact question shape — "Attempt at factorising their $9x^2-12x+4$" [real, June 2019, Q6(b) mark scheme, WMA12-verified-facts.md §9] — cited there only for its dM1 mark structure, never flagged for what the quotient itself actually is. Checked directly here: $9x^2-12x+4 = (3x-2)^2$ (expand back: $(3x-2)^2 = 9x^2 - 12x + 4$, confirmed). That question's factor was $(x-3)$, so its real, full factorisation is $f(x) = (x-3)(3x-2)^2$ — a genuine repeated linear factor, sitting inside an anchor this lesson already cites for an unrelated reason. This is the one row in this document's whole exam-angle ledger (below) that is a new independent finding rather than a re-statement of something the facts bank already flags as such.

#2b — The spec's own first named example, worked in full: a second repeated-factor instance

[original, illustrative, but the polynomial itself is the spec's own guidance text, not invented] $f(x) = x^3 + 3x^2 - 4$. Leading coefficient 1, constant $-4$: candidates are $x=\pm1,\pm2,\pm4$ only, smallest first. $f(1) = 1+3-4 = 0$ — $(x-1)$ is a factor on the very first try. Dividing (synthetic division, root 1, coefficients $1,3,0,-4$): $1,\; 1(1)+3=4,\; 4(1)+0=4,\; 4(1)-4=0$ — quotient $x^2+4x+4$. That quadratic is $(x+2)^2$ (expand back: $(x+2)^2 = x^2+4x+4$, confirmed), so $f(x) = (x-1)(x+2)^2$ — a second, independently confirmed repeated factor. The genuinely distinct point this sub-case adds to the trap-taxonomy: the lesson's existing answer-left-as-a-list-not-a- product trap [real, Jan 2020 mark scheme] covers writing three factors as a list instead of a product. A repeated factor invites a third, related error this document has not found named anywhere in this course's material — writing $(x-1)(x+2)(x+2)$ (a correct but non-standard product, listing the repeated bracket twice rather than with an index) instead of $(x-1)(x+2)^2$. This is reasoned directly from the algebra, not from a specific examiner-report finding, and is flagged as such rather than dressed up with an invented citation.

#2c — The spec's own second named example, worked in full: the non-monic full search

[original, illustrative; spec's own guidance polynomial] $f(x) = 6x^3 + 11x^2 - x - 6$. Here the leading coefficient is $6$, not $1$, so both halves of the constraint matter: any candidate root $\frac{b}{a}$ must have $b$ dividing $6$ ($\pm1,\pm2,\pm3,\pm6$) and $a$ dividing $6$ ($1,2,3,6$) — a candidate list of $\pm1, \pm2, \pm3, \pm6, \pm\frac12, \pm\frac13, \pm\frac16, \pm\frac23, \pm\frac32$, eighteen values, still finite. Testing smallest-magnitude first: $f(-1) = -6+11+1-6 = 0$ on the second try, so $(x+1)$ is a factor. Dividing (synthetic division, root $-1$, coefficients $6,11,-1,-6$): $6,\; 6(-1)+11=5,\; 5(-1)-1=-6,\; -6(-1)-6=0$ — quotient $6x^2+5x-6$. Factorising: need two numbers multiplying to $6\times(-6)=-36$ summing to $5$: $9$ and $-4$; split $6x^2+9x-4x-6 = 3x(2x+3)-2(2x+3) = (2x+3)(3x-2)$ (expand back: $(2x+3)(3x-2) = 6x^2+5x-6$, confirmed). So $f(x) = (x+1)(2x+3)(3x-2)$ — three distinct rational, non-integer-root factors, demonstrating the full non-monic search-and-divide-and-factor pipeline the lesson's own mechanism block sets up for this exact polynomial (calling the candidate list "a bigger list but still a list, not an infinity") but never carries through to a finished answer.

No real per-question WMA12 anchor has been found in the reviewed archive that requires factorising either 2b or 2c specifically. The spec names them only as guidance for the skill level expected, not as a claim they have appeared verbatim on a real paper — the lesson's own header states this explicitly, and this document repeats the same honesty rather than implying otherwise.

#2d — Boundary case: when the quotient does not factorise further at all

The lesson's teach block already states the rule: a quadratic quotient with a negative discriminant means the cubic simply has no third real linear factor, and stopping there is correct, not incomplete. Its own MCQ 6 demonstrates this once, on $f(x)=x^3+x^2-2x+12=(x+3)(x^2-2x+4)$, discriminant $4-16=-12$. A second, independent instance, with a simpler quadratic remainder: [original] $g(x) = x^3-2x^2+x-2$. $g(2) = 8-8+2-2=0$, so $(x-2)$ is a factor; dividing (synthetic division, root $2$, coefficients $1,-2,1,-2$): $1,\;1(2)-2=0,\;0(2)+1=1, \;1(2)-2=0$ — quotient $x^2+1$, discriminant $0-4=-4<0$. So $g(x) = (x-2)(x^2+1)$, and $x^2+1$ has no real root at all (not merely two irrational ones) — stopping here is not the factorisation-stopped- one-step-short trap, it is the only correct answer, because forcing a further real split that does not exist would itself be the error.

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#Part 3 — Sub-case: one condition, one unknown — a method-choice decision framework

The lesson's method-comparison block already gives the complete three-route, mark-by-mark treatment of this shape on $f(x)=2x^3+ax^2-7x+5$, remainder $3$ on division by $(x-2)$; it is not repeated here. One fresh, tiny instance to keep this section self-contained: [original] $f(x)=3x^3+ax^2+2x-5$, remainder $4$ on division by $(x+1)$. $f(-1) = -3+a-2-5 = a-10$; setting this to the stated remainder, $a-10=4$, gives $a=14$ — one substitution, one equation, one line, exactly the shape the facts bank's Jan 2024 Q1 citation describes as the paper's most efficient real route.

What no existing document states as a single explicit table is when each of the three credited methods is actually the right choice — synthesized here across the lesson's own three whenToPreferThis fields, the codex bank's method-choice note, and the facts bank's raw citations:

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#Part 4 — Sub-case: two independent conditions, two unknowns

The structural point neither the lesson nor the codex bank states explicitly. A cubic with two unknown coefficients needs exactly two independent linear equations to pin both down, because each condition — a stated factor, or a stated remainder — contributes exactly one linear equation once the identity is evaluated at that condition's own root. The lesson's marked-solution and the real Oct 2023 Q4 anchor both instantiate this the same way: one factor-condition paired with one remainder-condition. A pairing of two factor-conditions, or two remainder-conditions, would follow the identical mechanism — nothing about the theory rules it out — but no such pairing has actually been found in any series reviewed for this document, so it is stated here only as a structural possibility the same theory predicts, not as a confirmed real exam angle.

A fresh, fully worked instance of the confirmed (factor + remainder) pairing — distinct numbers from every worked example already in the lesson (worked-chain: $2x^3-5x^2-4x+3$/$(x-3)$; method-comparison: $2x^3+ax^2-7x+5$; marked-solution: $2x^3+px^2+qx+6$/$(2x-1)$/$(x+2)$):

[original] $f(x) = 2x^3+px^2+qx-3$. Given $(x-1)$ is a factor and the remainder on division by $(x+2)$ is $9$, find $p$, $q$, and hence factorise $f(x)$ completely.

• $(x-1)$ a factor $\Rightarrow f(1)=0$: $2+p+q-3=0 \Rightarrow p+q=1$.
• Remainder $9$ on division by $(x+2)$ $\Rightarrow f(-2)=9$: $-16+4p-2q-3=9 \Rightarrow 4p-2q=28 \Rightarrow 2p-q=14$.
• Adding the two equations: $3p=15 \Rightarrow p=5$; then $q=1-5=-4$.
• So $f(x) = 2x^3+5x^2-4x-3$. Check: $f(1)=2+5-4-3=0$ ✓; $f(-2)=-16+20+8-3=9$ ✓.
• Divide by $(x-1)$ (synthetic division, root $1$, coefficients $2,5,-4,-3$): $2,\;2(1)+5=7, \;7(1)-4=3,\;3(1)-3=0$ — quotient $2x^2+7x+3$.
• Factorise: need two numbers multiplying to $2\times3=6$ summing to $7$: $6$ and $1$; $2x^2+6x+x+3 = 2x(x+3)+1(x+3)=(2x+1)(x+3)$ (expand back: confirmed, $2x^2+7x+3$).
• $f(x) = (x-1)(2x+1)(x+3)$.

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#Part 5 — What's already covered, cross-referenced rather than restated

The lesson's eight trap-taxonomy items already carry every real, cited error pattern this document has found for the single-condition and given-factor shapes (Parts 1 and 3 above): the root-sign error, the remainder-set-to-zero error, division chosen over substitution, factorisation stopped one step short, "hence" ignored, a non-integer factor form, an answer left as a list, and missing "show that" steps — each with its own real citation, not repeated here. This section adds only the one new, uncited notational point Part 2b surfaces (repeated-factor notation) and cross-references where the rest already live, so a reader working through this document knows exactly where each sub-case's error patterns are actually taught: single-condition and given-factor traps in the lesson's own trap-taxonomy block; the search-from-scratch sub-case (Part 2 above) currently has none named anywhere in this course, beyond the reasoned repeated-factor-notation point above.

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#Part 6 — Why this skill recurs outside its own spec item

Every real WMA12 sighting of a cubic being factorised sits inside spec 2.1 itself. But the same underlying skill — testing values, factorising a polynomial, and then keeping the exact form of a non-integer root rather than rounding it — recurs as a step inside questions whose main spec point is differentiation, confirmed in two independent lessons' own audit trails and pulled together here for the first time under this topic:

Jan 2021, Q2 [real, stationary-points-and-curve-sketching.ts audit, WMA12-verified-facts.md §9]: for $y=x^3-x^2-16x+2$, $\frac{dy}{dx}=3x^2-2x-16$. Factorising the quadratic derivative (the same factor-testing skill, one degree down): $3x^2-2x-16=(3x-8)(x+2)$ (expand back: confirmed), giving critical $x$-values $\frac83$ and $-2$. The real mark scheme's own accuracy line requires the fraction kept exact — "do not accept 2.67" — the identical "keep it exact, don't round" discipline Part 2 above and the lesson's own show-that-steps-omitted trap both rely on, independently confirmed on a completely different spec strand.

June 2025, WMA12/01A, Q8 [real, WMA12-verified-facts.md §6.2 and §9's corrected paper-code entry — corrected here from the lesson's own trap-taxonomy citation, which still labels this series "Regional 01R"; the facts bank's own later audit confirms the real paper's cover reads "WMA12/01A"]: part (a) gives the derivative already factorised, $f'(x)=2(3x-2)(x+5)$, to find where a curve is decreasing (critical values $\frac23$ and $-5$, interval $-5<x<\frac23$; "either of these...scores both marks"). Part (b) then needed a linear factor in integer-coefficient form; a real candidate wrote it as $x-\frac72$ instead of $(2x-7)$ — "this is incorrect as integer values were required" — the identical non-integer-factor-form-given trap the lesson's own trap-taxonomy already names, confirmed here as a genuinely cross-topic sighting rather than a spec-2.1-only occurrence.

The pattern across both rows: every confirmed cross-topic sighting of this skill found in the reviewed archive comes from a differentiation/curve-sketching question, where finding a stationary point or a sign-change interval requires factorising a derivative — never from an integration or circle question. That is a real, if narrow, observation from two rows, not a claim the skill is absent elsewhere; no integration- or circle-context sighting was searched for and found in this pass.

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#Every real exam angle this topic has actually been tested from — consolidated ledger

Read every row as "seen in the series named," never as a probability or a frequency claim beyond what is stated — this document invents no likelihood percentage and no grade boundary anywhere, per this course's standing discipline.

Cross-cutting observations, each traceable to a specific row above, none invented:

• This topic has a genuine real repeated-factor anchor (row 3), independently identified in this document by checking whether an already-cited real quotient happens to be a perfect square — it is, and no existing course document had flagged it.
• The two "hence" and "non-integer factor form" traps are each confirmed on two structurally different question types — one dedicated (row 2), one embedded in a differentiation question (row 7) — real evidence these are general marking disciplines attached to the skill, not quirks of one question.
• The "keep it exact" discipline recurs on a completely different spec strand (row 8), confirming it is a paper-wide convention rather than something specific to remainder/factor questions.
• No real per-question anchor exists for either of the spec's own two named example cubics (Part 2b, 2c above) as tested verbatim — they are worked here because the spec names them as the expected skill level, not because either has been found on a real paper.
• No grade-boundary document and no likelihood figure for this topic specifically appear in anything reviewed for this document; none is stated here for that reason.

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The WMA11 prerequisite this lesson depends on — factorising the quadratic quotient itself, and checking its discriminant before assuming a further real factor exists — already has its own dedicated coverage in wma11/quadratic-functions.ts and is not duplicated here; Part 2d above uses it only as the boundary case for this topic's own final step.