Algebraic Division, the Factor Theorem and the Remainder Theorem

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

What spec 2.1 actually asks for — and what the formula booklet does not give you

The whole of P2's "Algebra and functions" content area is one dotted item, 2.1: "Simple algebraic division; use of the Factor Theorem and the Remainder Theorem." Its guidance narrows the scope usefully: "Only division by $(ax + b)$ or $(ax - b)$ will be required... Students may be required to factorise cubic expressions such as $x^3 + 3x^2 - 4$ and $6x^3 + 11x^2 - x - 6$ ." Two things follow immediately. Every divisor you will meet is linear — a single power of $x$ , possibly with a coefficient in front — so the degree bookkeeping never gets complicated. And the target skill is not division for its own sake: it is factorising a cubic completely, with division as one of the tools that gets you there.

Now the part that decides how you revise this topic. This course's own audit of the *Mathematical Formulae and Statistical Tables* booklet — the yellow booklet you are given in the exam — records that on the P2 page, "Proof, algebraic division/factor theorem, circle geometry, and differentiation/integration technique... have no formulae listed at all". There is nothing to look up here. Arithmetic and geometric series formulae are printed; the is printed; the is printed. The remainder theorem is not, the factor theorem is not, and no division layout is. That is not a reason to memorise two more statements — it is a reason to be able to *rebuild* both from the one identity the mechanism block below derives, which takes one line and cannot be forgotten in the way an unmotivated formula can.

Questions on this item come in two shapes, and telling them apart is the first decision to make. Shape one: a remainder is given or asked for. "Find the remainder when $f(x)$ is divided by $(x-3)$ ", or, more often, " $f(x)$ leaves remainder $7$ when divided by $(x-3)$ ; find the constant $a$ ." Shape two: a factor is given or asked for. "Show that $(2x-1)$ is a factor", or "factorise $f(x)$ completely." Shape two is shape one with the remainder set to zero — but the marks are earned differently, because a factor question almost always continues into a division and a full factorisation, while a remainder question usually stops at one substitution. The commonest conceptual error on the whole item is running the two shapes together: a real examiner report on an accessible opening question records candidates "setting equal to $-3$ or $0$ , rather than $3$ " — that is, forcing a stated remainder to be zero because factor-theorem questions are the ones they had practised (Jan 2024, Q1).

Mechanism

Where both theorems come from — one identity, evaluated at the value that deletes the quotient

Start with what division produces, before either theorem is stated. Dividing $f(x)$ by a linear expression $(ax - b)$ gives some quotient $q(x)$ and leaves some remainder $r$ , and that is exactly the statement

$f(x) \equiv (ax - b)\,q(x) + r.$

Two details in that line are doing all the work later, so they are worth pinning down now. First, $r$ is a constant. The division only stops when what is left over is too small to be divided again — if anything of degree 1 or more remained, the divisor (degree 1) would still go into it — so the remainder's degree must be strictly less than the divisor's, and "less than degree 1" means degree 0: a plain number, possibly zero. Second, the symbol is $\equiv$ , not $=$ . This is an identity: the two sides are the same expression written two ways, so it is true for every single value of $x$ , not for a special few. That is not pedantry — it is the entire lever.

Because the identity holds for every $x$ , you are free to choose the most convenient $x$ there is. Choose the value that makes the divisor zero. $(ax - b) = 0$ when $x = \frac{b}{a}$ , so substituting that value gives

$f\!\left(\tfrac{b}{a}\right) = \left(a \cdot \tfrac{b}{a} - b\right) q\!\left(\tfrac{b}{a}\right) + r = 0 \times q\!\left(\tfrac{b}{a}\right) + r = r.$

That is the remainder theorem, and notice what it cost: nothing was assumed about $q(x)$ at all. Whatever the quotient is — however long, however awkward — it is multiplied by zero and disappears. This is why substitution beats division for a remainder: the division's entire output, the quotient, is the part you were never asked for, and choosing $x = \frac{b}{a}$ deletes it in one step rather than computing it and then throwing it away.

The factor theorem is now not a second fact. Saying $(ax-b)$ is a factor of $f(x)$ means it divides exactly — i.e. $r = 0$ — and the identity then reads $f(x) \equiv (ax-b)q(x)$ . Run the same substitution: $f\!\left(\frac{b}{a}\right) = r = 0$ . And the converse runs backwards down the identical line: if $f\!\left(\frac{b}{a}\right) = 0$ then $r = 0$ , so $f(x) \equiv (ax-b)q(x)$ and $(ax-b)$ is a factor. Both directions come from the one identity, which is why the factor theorem is an "if and only if" rather than a one-way test: $(ax - b)$ is a factor of $f(x)$ exactly when $f\!\left(\frac{b}{a}\right) = 0$ .

Everything the exam does to trip you here is a distortion of the substitution step, so read it once more slowly. The value substituted is the value that makes the *divisor* zero — solve $ax - b = 0$ , do not read a number off the bracket. For $(x - 2)$ that value is $+2$ ; for $(x + 2)$ it is $-2$ ; for $(2x + 1)$ it is $-\frac{1}{2}$ , not $\frac{1}{2}$ and not $-2$ . And what the substitution equals is $r$ , the *stated remainder* — which is zero only when the question has actually said "factor".

Doing the division itself: long division, or writing the identity down and comparing coefficients

Once a factor is known, a "factorise completely" question still needs the quotient, and there the remainder theorem has nothing left to offer — the quotient is precisely the thing substitution deletes. Two methods produce it, and the spec's "simple algebraic division" wording covers both. Algebraic long division is the familiar layout: divide the leading term, multiply back, subtract, bring down, repeat. It works, it is fully creditable, and its weakness is mechanical rather than conceptual — every step involves subtracting a bracket, and each of those subtractions is a fresh opportunity for a sign error to enter working that then continues confidently downwards.

Comparing coefficients is the same identity from the mechanism block, used as a working tool instead of a proof. Degrees are forced: dividing a cubic by a linear expression must leave a quadratic, because the degrees add when you multiply back ( $3 = 1 + 2$ ). So for a cubic $f(x)$ with known factor $(x - k)$ you can write down, immediately, $f(x) \equiv (x-k)(Ax^2 + Bx + C)$ with $A$ , $B$ , $C$ unknown, expand the right-hand side once, and match coefficients term by term. The leading coefficient and the constant term usually fall out instantly — the $x^3$ coefficient gives $A$ with no work at all, and the constant gives $C$ from a single small equation — leaving only the middle coefficient to pin down, which then acts as a free check because it must agree. Nothing is ever subtracted, so the sign errors long division generates have nowhere to enter.

Neither method is "the" method. The general marking guidance sets out method marks for the standard techniques and credits a correct approach carried out correctly, and the method-comparison block below shows three routes to the same answer for the same marks. What differs is the failure rate under time pressure, which is a real quantity the examiners have measured and reported, not a matter of taste.

The final step of a full factorisation belongs to the WMA11 prerequisite rather than to this item: once the quadratic quotient is in hand, it is factorised the way any quadratic is, and if it does not factorise — negative discriminant — then the cubic simply has no third real linear factor and the correct final answer keeps the quadratic intact. Checking the discriminant before hunting for factors that do not exist is the same habit the pilot lesson builds for quadratics, applied one level up.

Mechanism

Which values are worth trying — why the constant term tells you where to look, instead of guessing

A question that gives you a factor has done the hard part. A question that says only "factorise $f(x)$ completely" has not, and trying values at random is both slow and unbounded. The identity constrains the search hard enough that you rarely need more than two or three trials, and the constraint is derivable rather than magical.

Suppose $f(x)$ has integer coefficients, and suppose it factorises as $f(x) \equiv (ax - b)q(x)$ where $q(x)$ also has integer coefficients — which is the only kind of factorisation this spec's own guidance asks for, since it names cubics like $6x^3 + 11x^2 - x - 6$ that split into integer-coefficient linear factors. Now look at just two coefficients of that product. The constant term of $f$ is what you get by multiplying the two constant terms together: $-b \times (\text{constant term of } q)$ . Since everything in sight is an integer, $b$ must divide $f$ 's constant term. The leading coefficient of $f$ is likewise the product of the two leading coefficients: $a \times (\text{leading coefficient of } q)$ , so $a$ must divide $f$ 's leading coefficient.

That is the whole search space. For a cubic like $x^3 + x^2 - 2x + 12$ , the leading coefficient is 1, so $a = 1$ and the only candidates are $x - b$ with $b$ dividing 12: try $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$ and nothing else — and start with the small ones, because they are both likelier and quicker to evaluate. For a cubic like $6x^3 + 11x^2 - x - 6$ , $a$ must divide 6 and $b$ must divide 6, which is a bigger list but still a list, not an infinity.

Two consequences worth carrying into an exam. First, if every candidate value fails, that is information, not a dead end: the cubic has no linear factor with integer coefficients, and the question was never asking for one. Second, this is why factor-theorem questions almost always hand you a small integer or a simple fraction like $\frac{1}{2}$ or $-\frac{2}{3}$ — the setters are working from the same constraint, in reverse.

Worked, in full

Given that $(x - 3)$ is a factor of $f(x) = 2x^3 - 5x^2 - 4x + 3$ , factorise $f(x)$ completely — VERIDIAN-original

01
Confirm the given factor rather than assuming it, since it takes one line and secures the ground everything else is built on. $(x-3)$ is zero at $x = 3$ , so evaluate $f(3) = 2(27) - 5(9) - 4(3) + 3 = 54 - 45 - 12 + 3 = 0$ . By the factor theorem, a zero value at the divisor's root is exactly what "is a factor" means, so $(x-3)$ divides $f(x)$ exactly.
Earns: Not separately scored — good exam practice rather than a marked step. Two real mark schemes for this exact question shape (a factor already established, now divide it out and factorise completely) confirm this directly: both start their marking at the division itself, with no antecedent mark for re-confirming a factor the question has already stated as given. "M1: Attempt to divide or factorise out (x − 3)" is the very first mark in one (June 2019, Q6(b)); the other begins identically, "M1: Attempt to divide or factorise out (2x + 1)" (Oct 2023, Q4(c)(ii)). Writing this check anyway is still worth doing — it catches a misread question before any of the marked working starts — but it does not itself earn marks here.
02
Set up the division by degree bookkeeping instead of reaching for a layout. A cubic divided by a linear expression must leave a quadratic, so write the identity with unknown coefficients: $2x^3 - 5x^2 - 4x + 3 \equiv (x - 3)(2x^2 + bx + c)$ . The leading coefficient is already forced — $2x^2 \times x = 2x^3$ — so only $b$ and $c$ remain.
Earns: M1 — method mark for a complete and correct division strategy. Long division of $f(x)$ by $(x-3)$ earns the identical mark; the general marking guidance credits a correct method carried out correctly, not one particular layout.
03
Expand the right-hand side and compare coefficients: $(x-3)(2x^2 + bx + c) = 2x^3 + bx^2 + cx - 6x^2 - 3bx - 3c = 2x^3 + (b-6)x^2 + (c-3b)x - 3c$ . The constant term gives $-3c = 3$ , so $c = -1$ . The $x$ term gives $c - 3b = -4$ , so $-1 - 3b = -4$ and $b = 1$ . The $x^2$ term is now a free check: $b - 6 = 1 - 6 = -5$ ✓, matching $f$ 's own $-5x^2$ . The quotient is $2x^2 + x - 1$ .
Earns: A1 — the correct quadratic quotient. Note the built-in check: the third coefficient was not needed to find b and c, so its agreement independently confirms the other two rather than merely repeating them.
04
Factorise the quadratic quotient, which is WMA11 work rather than anything new here: $2x^2 + x - 1 = (2x - 1)(x + 1)$ . Expanding back confirms it: $2x^2 + 2x - x - 1 = 2x^2 + x - 1$ ✓.
Earns: dM1 — method mark for a valid attempt at factorising the three-term quadratic, DEPENDENT on the division above (you cannot factorise a quotient you have not yet correctly found). Two real mark schemes for this exact question shape tag this identical step dependent, not independent: "dM1: Attempts to factorise their $3x^2+4x-4$ " (Jan 2020, Q3(b)) and "dM1: Attempt at factorising their $9x^2-12x+4$ . Apply the usual rules for factorising" (June 2019, Q6(b)) — both explicitly "their" quotient, i.e. dependent on the preceding method-and-accuracy pair having produced it.
05
Write the complete factorisation as a single product of three linear factors: $f(x) = (x - 3)(2x - 1)(x + 1)$ . Nothing further will factorise — all three factors are linear — so this is what "completely" asks for.
Earns: A1 — the fully factorised answer, written as a product. A real mark scheme records the required form explicitly, and a solution that stops at $(x-3)(2x^2+x-1)$ has not answered the question asked.

Source — Mark scheme, Jan 2020

"(2x+3)(3x−2)(x+2) written out as a product of factors (not as a list) and isw"

Complete it yourself

Complete the chain — show that $(x + 2)$ is a factor of $f(x) = 3x^3 + 2x^2 - 7x + 2$ , and factorise $f(x)$ completely

01
Find the value that makes the divisor zero, taking care with the sign: $(x + 2) = 0$ gives $x = -2$ , not $x = 2$ . That is the value the factor theorem needs, and reading $+2$ straight off the bracket is the single most-documented error on this spec item.
02
Evaluate there, showing the substitution rather than only the result: $f(-2) = 3(-8) + 2(4) - 7(-2) + 2 = -24 + 8 + 14 + 2 = 0$ . Since $f(-2) = 0$ , the factor theorem gives that $(x+2)$ is a factor of $f(x)$ .

Same question, every valid method

$f(x) = 2x^3 + ax^2 - 7x + 5$ , where $a$ is a constant. When $f(x)$ is divided by $(x - 2)$ the remainder is $3$ . Find the value of $a$ . (3 marks) — VERIDIAN-original question, in the shape the facts bank confirms is this topic's standard accessible opener; not a reproduction of any past-paper question.

3 valid methods · every one reaches $a = -1$ · 3 marks available

01
$(x-2)$ is zero at $x = 2$ , so by the remainder theorem the remainder equals $f(2)$ : $f(2) = 2(2)^3 + a(2)^2 - 7(2) + 5 = 16 + 4a - 14 + 5 = 4a + 7$ . The remainder is given as $3$ , so $4a + 7 = 3$ .
Method mark for substituting the value that makes the divisor zero into f(x) and setting the result equal to the stated remainder. The mark is for the method being evident, so it survives an embedding or sign slip in reaching the equation — matching the verified anchor's own wording for this exact step: "Attempts to set f(2) = 3" (Jan 2024, Q1).
M1
02
$4a = 3 - 7 = -4$ .
Method mark, dependent on the first, for the solving step: isolating the unknown from the equation just formed. Tolerant of an arithmetic or rearrangement slip provided the method is evident — the verified anchor treats this as a second, independent method mark rather than folding it into the accuracy mark: "Solves a linear equation in a arising from setting f(2) = 3. Condone slips in their rearrangement proceeding to a = ..." (Jan 2024, Q1).
M1
03
$a = -1$ .
Accuracy mark, cao — the bare final numeric value only, nothing else bundled into this line, matching the verified anchor's own final line: "A1: a = 5/8 or exact equivalent" (Jan 2024, Q1) — the value differs because the question does, the mark structure does not.
A1

Almost always, when a remainder or an unknown coefficient is what the question wants. It is three lines, it never subtracts a bracket, and the examiner record on a real question of exactly this shape is unusually direct: "The two main approaches were either using the remainder theorem or using algebraic long division. The remainder theorem was the most efficient approach, and this was also the most common" (Jan 2024, Q1). Show the substitution written out, not just its value — the general marking guidance's advice is that a learnt method should be quoted before it is used, since an implied method can still be credited but "may be lost if there is any mistake in the working".

Marked, line by line

$f(x) = 2x^3 + px^2 + qx + 6$ , where $p$ and $q$ are constants. (a) Given that $(2x - 1)$ is a factor of $f(x)$ , show that $p + 2q = -25$ . (2) (b) Given also that when $f(x)$ is divided by $(x + 2)$ the remainder is $20$ , show that $2p - q = 15$ . (2) (c) Find the value of $p$ and the value of $q$ , and hence factorise $f(x)$ completely. (5) — VERIDIAN-original question, in the three-part shape the facts bank confirms this spec item is set in (a "show that" from a given factor, then simultaneous equations feeding a full factorisation — Oct 2023 Q4). This version makes part (b) a second "show that" too, rather than the real anchor's un-printed "find a second simplified equation" — a deliberate original variant, not the real question's own structure: it lets part (c) demonstrate recovering with a printed equation even after a part (b) slip, which the real anchor's own unprinted equation cannot do. Not a reproduction of any past-paper question: neither the polynomial nor either printed equation appears in the facts bank's transcription of any real series, and the per-line mark allocations are modelled on verified mark-scheme conventions (M/A/B definitions, cso and ag on printed answers, cao by default on A marks) rather than copied from a real scheme, which for an original question does not exist.

9 marks available

(a) — 2 marks

01
$(2x-1) = 0$ when $x = \frac{1}{2}$ , and $(2x-1)$ being a factor means $f\left(\frac{1}{2}\right) = 0$ . Substituting: $2\left(\frac{1}{8}\right) + p\left(\frac{1}{4}\right) + q\left(\frac{1}{2}\right) + 6 = 0$ .
Method mark for setting f(½) = 0 — the value that makes the given factor zero, with the substitution written out rather than only its result.
M1
02
Multiply every term by $4$ to clear the fractions: $1 + p + 2q + 24 = 0$ , so $p + 2q = -25$ , as required.
Accuracy mark, cso — the answer is printed on the paper (ag), so there must be no errors in this part to earn it. The intermediate line clearing the fractions is what makes the printed result follow visibly; a solution jumping from the substitution straight to the printed answer has shown nothing.
A1

(b) — 2 marks

101
$(x+2) = 0$ when $x = -2$ , so by the remainder theorem the remainder is $f(-2)$ , and the question states this equals $20$ : $2(-8) + p(4) + q(-2) + 6 = 20$ .
Method mark for setting f(−2) equal to the stated remainder. Two decisions are being marked at once: the sign of the value substituted, and setting the result to 20 rather than to 0.
M1
102
$-16 + 4p - 2q + 6 = 20$ , so $4p - 2q = 30$ , and halving gives $2p - q = 15$ , as required.
Accuracy mark, cso — again a printed answer (ag), so every intermediate line has to be present and correct.
A1

(c) — 5 marks

201
Solve the two printed equations simultaneously. From $2p - q = 15$ , $q = 2p - 15$ . Substituting into $p + 2q = -25$ : $p + 2(2p - 15) = -25$ , so $5p - 30 = -25$ and $5p = 5$ .
Method mark for a complete method of solving the simultaneous equations — substitution or elimination, either is credited.
M1
202
$p = 1$ , and then $q = 2(1) - 15 = -13$ . So $f(x) = 2x^3 + x^2 - 13x + 6$ .
Accuracy mark for both values. Worth a five-second check against part (a): 1 + 2(−13) = −25 ✓, and against (b): 2(1) − (−13) = 15 ✓.
A1
203
Divide by the factor already known from (a). A cubic over a linear expression leaves a quadratic, so $2x^3 + x^2 - 13x + 6 \equiv (2x-1)(x^2 + bx + c)$ ; the $x^3$ term forces the leading coefficient $1$ , the constant gives $-c = 6 \Rightarrow c = -6$ , and the $x$ term gives $-b + 2c = -13 \Rightarrow -b - 12 = -13 \Rightarrow b = 1$ .
Method mark for a full attempt at dividing f(x) by the known factor (2x − 1), by long division or by comparing coefficients — the scheme credits either.
M1
204
The quotient is $x^2 + x - 6$ . Check with the $x^2$ coefficient, which was not used: $2b - 1 = 2(1) - 1 = 1$ ✓.
Accuracy mark for the correct quadratic quotient.
A1
205
Factorise the quotient: $x^2 + x - 6 = (x+3)(x-2)$ . So $f(x) = (2x - 1)(x + 3)(x - 2)$ .
Accuracy mark for the completely factorised cubic, written as a product of three linear factors with integer coefficients — the form a real mark scheme specifies, adding that once it is written out as a product, subsequent working is ignored (isw).
A1

In your own words

In one sentence: why does substituting $x = \frac{b}{a}$ into $f(x)$ give the remainder on dividing by $(ax - b)$ , without your ever having to know what the quotient is?

Named traps

root-sign-taken-straight-from-the-bracket: The dominant error on this spec item, confirmed on a real three-part factor-and-remainder question in both directions at once: candidates were recorded "setting $f(\frac{1}{2}) = 0$ in part (a) and/or setting $f(-2) = -25$ in part (b)" where the correct substitutions were $f(-\frac{1}{2})$ and $f(2)$ (Oct 2023, Q4). The fix is mechanical rather than mnemonic: never read a number off a bracket, solve the bracket for $x$ . $(2x+1)=0$ gives $x = -\frac{1}{2}$ ; $(x+2)=0$ gives $x = -2$ ; $(3x-2)=0$ gives $x = \frac{2}{3}$ . One extra line, and the most expensive error on the topic cannot happen.
remainder-equation-set-to-zero: Confirmed on an accessible opening question: the commonest conceptual errors were "using $f(-2)$ or $f(3)$ rather than $f(2)$ or setting equal to $-3$ or $0$ , rather than $3$ " (Jan 2024, Q1). Setting the substitution equal to zero is the factor-theorem reflex applied to a remainder question — and it is worth seeing that it is the same identity in both cases, $f(x) \equiv (ax-b)q(x) + r$ , with $r$ being whatever the question said it was. Zero is a value of $r$ , not the default.
long-division-chosen-when-substitution-was-asked-for: Not wrong, but measurably worse, and the examiners have said so in as many words: "Candidates who chose to use division gave themselves a more difficult task and tended to be less successful, with many giving up part way through the division" (Jan 2024, Q1), on a question where "the remainder theorem was the most efficient approach." Division computes the entire quotient in order to discard it. Reach for it when a later part actually wants the quotient; otherwise substitute.
factorisation-stopped-one-step-short: Confirmed as a specific, separately-recorded cause of lost marks: "leaving the answer to (c) as $(2x+1)(2x^2 + x - 15)$ " (Oct 2023, Q4). That quadratic does factorise — $2x^2 + x - 15 = (2x-5)(x+3)$ — which is exactly why the answer was marked short: "factorise completely" is not satisfied while any factor still factorises. After dividing out the known factor, always ask whether the quadratic quotient breaks down further, and check its discriminant if it is not obvious.
hence-factorise-reached-without-using-the-given-factor: On a "hence factorise" part, the word "hence" is a mark-scheme instruction, not decoration: the working must derive the factorisation from the already-established factor, and reaching the identical, completely correct final answer by any other route earns nothing on that part. A real mark scheme is explicit about it, on the same question this lesson's worked chain and marked solution are modelled on: "The question states 'Hence' so there is an expectation that they use the factor of $(2x+1)$ which is given. Solutions that just state $4x^3 + 4x^2 - 29x - 15 = (2x+1)(2x-5)(x+3)$ score M0 and therefore A0 A0" (Oct 2023, Q4(c)(ii)) — a fully correct factorisation, worth zero, purely because the shown working never used the given factor. The matching real error is recorded in the same series' examiner report: candidates "using a calculator to solve $4x^3 + 4x^2 - 29x - 15 = 0$ and writing $x = -3, -\frac{1}{2}, \frac{5}{2}$ " — solving the cubic directly, on a paper where calculators are permitted throughout, instead of dividing out the factor the question had already given them. Every worked chain, chain drill and marked solution in this lesson models the division route for exactly this reason: on a "hence" question, showing the division or comparison is not supporting style, it is the entire mark allocation for that part, independent of whether the final product is right.
non-integer-factor-form-given: A genuine "mathematically equivalent but marked wrong" trap, confirmed on a real question whose final step was a factorisation: a candidate "giving the factor as $x - \frac{7}{2}$ ... This is incorrect as integer values were required" (June 2025 Regional 01R, Q8). $\left(x - \frac{7}{2}\right)$ and $(2x-7)$ have the same root and are the same factor up to a constant, and the scheme still wants the integer-coefficient form. When a root comes out as a fraction $\frac{b}{a}$ , write the factor as $(ax - b)$ .
answer-left-as-a-list-not-a-product: A real mark scheme states the required form of a factorised cubic explicitly, requiring it "written out as a product of factors (not as a list) and isw" (Jan 2020). Listing the three factors, or listing the three roots, is not the same object as their product — the question asked for $f(x)$ factorised, so the answer is an expression equal to $f(x)$ . The isw (ignore subsequent working) note is the encouraging half: once the product is correctly written down, later scribbling does not undo it.
show-that-steps-omitted: On a "show that" part, the printed answer is not the thing being marked — the route to it is. A real report records marks lost for "not showing the $= 0$ or intermediate lines when attempting to prove (a)" (Oct 2023, Q4), and the general marking guidance's cso convention makes the standard explicit: "There must be no errors in this part of the question to obtain this mark." Write the substitution, write the line that clears the fractions, then write the printed result — a solution that leaps from one to the other has demonstrated nothing that could be checked.

Retrieval — with feedback on every choice

Question 1

2 marks

$f(x) = 4x^3 - 4x^2 + 3x + 5$ . Find the remainder when $f(x)$ is divided by $(2x - 1)$ .

Question 2

2 marks

$f(x) = x^3 + kx^2 + 3x + 9$ , where $k$ is a constant. Given that $(x + 3)$ is a factor of $f(x)$ , find the value of $k$ .

Question 3

3 marks

$f(x) = 2x^3 + 3x^2 - 8x + 3$ , and $(x-1)$ is a factor. Which is $f(x)$ written as a product of three linear factors with integer coefficients?

Question 4

2 marks

$f(x) = x^3 + ax^2 + 5x + 4$ , where $a$ is a constant. When $f(x)$ is divided by $(x - 3)$ the remainder is $10$ . Which equation correctly captures that information?

Question 5

1 mark

A question gives a cubic with one unknown coefficient and states the remainder on division by a linear expression, then asks for that coefficient. What do the examiners' own records say about the two available routes?

Question 6

2 marks

$f(x) = x^3 + x^2 - 2x + 12$ . Which of these is a factor of $f(x)$ ?

Reference — not a study method, a lookup

Division identity: f(x) ≡ (ax − b)q(x) + r; r is constant for a linear divisor.
Remainder theorem: dividing by (ax − b) leaves f(b/a); by (ax + b), f(−b/a). Solve the bracket, never read it off.
Factor theorem: (ax − b) is a factor ⟺ f(b/a) = 0 — the same identity with r = 0.
Set f(root) equal to the stated remainder, not 0, unless told 'factor'.
Factorise completely: divide out the known factor, factorise the quadratic quotient, answer as a product of integer-coefficient linear factors.
Remainder wanted? Substitute. Quotient wanted? Divide, or compare coefficients.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation and figure attributed to a mark scheme or examiner report in this lesson was independently verified against the primary Pearson document, not carried over from prior course material. Every question in this lesson — prequestion, worked chain, chain drill, method comparison, marked solution and MCQ alike — is VERIDIAN-original wording, inspired by confirmed real question types, never a reproduction of a real Pearson question; and because the questions are original, the per-line mark allocations attached to them are modelled on verified mark-scheme conventions (what M, A and B marks mean, what cso and ag require on a printed answer, when isw applies) rather than transcribed from a real mark scheme, which for an original question does not exist.

Question 12 marks

$f(x) = 4x^3 - 4x^2 + 3x + 5$ . Find the remainder when $f(x)$ is divided by $(2x - 1)$ .

$6$
Correct. $(2x-1) = 0$ at $x = \frac{1}{2}$ , so the remainder is $f\left(\frac{1}{2}\right) = 4\left(\frac{1}{8}\right) - 4\left(\frac{1}{4}\right) + \frac{3}{2} + 5 = 0.5 - 1 + 1.5 + 5 = 6$ .
B $2$
This is $f\left(-\frac{1}{2}\right) = -0.5 - 1 - 1.5 + 5 = 2$ . The sign has been taken from the bracket rather than found by solving it: $2x - 1 = 0$ gives $x = +\frac{1}{2}$ .
C $27$
This is $f(2) = 32 - 16 + 6 + 5 = 27$ , using the coefficient of $x$ as though it were the root. In $(2x-1)$ the $2$ multiplies $x$ ; the value that makes the divisor zero is $\frac{1}{2}$ , its reciprocal, not $2$ .
D $12$
This doubles the correct remainder, on the assumption that dividing by $(2x-1)$ rather than $\left(x - \frac{1}{2}\right)$ scales what is left over. It scales the quotient, not the remainder: $f(x) \equiv (2x-1)q(x) + r$ and $f(x) \equiv \left(x-\frac{1}{2}\right)\left(2q(x)\right) + r$ are the same identity, with the same $r$ . Here the quotients are $2x^2 - x + 1$ and $4x^2 - 2x + 2$ — genuinely double — while the remainder is $6$ either way.

Traps tested: Root sign taken straight from the bracket · Root read as the coefficient · Remainder scaled with the divisor

Question 22 marks

$f(x) = x^3 + kx^2 + 3x + 9$ , where $k$ is a constant. Given that $(x + 3)$ is a factor of $f(x)$ , find the value of $k$ .

$k = 3$
Correct. $(x+3) = 0$ at $x = -3$ , so the factor theorem gives $f(-3) = 0$ : $(-27) + 9k - 9 + 9 = 9k - 27 = 0$ , hence $k = 3$ . (Checking: $x^3 + 3x^2 + 3x + 9 = (x+3)(x^2+3)$ , and $x^2+3$ has no real roots, so the factorisation stops there.)
B $k = -5$
This substitutes $x = +3$ : $27 + 9k + 9 + 9 = 9k + 45 = 0$ gives $k = -5$ . The value has been read off the bracket instead of found from it — $(x+3)$ is zero at $x = -3$ .
C $k = -3$
This substitutes $-3$ but treats $(-3)^3$ as $+27$ : $27 + 9k - 9 + 9 = 9k + 27 = 0$ gives $-3$ . An odd power of a negative number stays negative — $(-3)^3 = -27$ — while the even power in the $kx^2$ term does turn positive, which is why the two terms behave differently in the same substitution.
D $k = 9$
This treats the $kx^2$ term as $3k$ rather than $9k$ : $-27 + 3k - 9 + 9 = 3k - 27 = 0$ gives $9$ . The substituted value has to be squared before it multiplies $k$ , since the term is $kx^2$ and $(-3)^2 = 9$ .

Traps tested: Root sign taken straight from the bracket · Odd power sign error · Square not applied to substituted value

Question 33 marks

$f(x) = 2x^3 + 3x^2 - 8x + 3$ , and $(x-1)$ is a factor. Which is $f(x)$ written as a product of three linear factors with integer coefficients?

$(x-1)(2x-1)(x+3)$
Correct. Dividing by $(x-1)$ gives the quotient $2x^2 + 5x - 3$ , which factorises as $(2x-1)(x+3)$ . Multiplying back confirms it: $(2x-1)(x+3) = 2x^2 + 5x - 3$ , and $(x-1)(2x^2+5x-3) = 2x^3 + 3x^2 - 8x + 3$ .
B $(x-1)(2x^2 + 5x - 3)$
The division is right and this expression does equal $f(x)$ — but the quadratic still factorises, so this is one step short of the answer asked for. A real report records exactly this loss, an answer "left... as $(2x+1)(2x^2 + x - 15)$ " on a factorise-completely question (Oct 2023, Q4).
C $2(x-1)\left(x - \tfrac{1}{2}\right)(x+3)$
Mathematically this equals $f(x)$ — $2\left(x-\frac{1}{2}\right) = 2x-1$ — and it still is not the required form. A real report records a candidate "giving the factor as $x - \frac{7}{2}$ ... This is incorrect as integer values were required" (June 2025 Regional 01R, Q8): a fractional root $\frac{b}{a}$ is written as the factor $(ax-b)$ .
D $(x-1)(2x-1)(x-3)$
The signs in the last factor are reversed. Check by expanding: $(2x-1)(x-3) = 2x^2 - 7x + 3$ , not the quotient $2x^2 + 5x - 3$ the division produced. Multiplying the factors back out catches this in one line.

Traps tested: Factorisation stopped one step short · Non integer factor form given · Quadratic factor signs reversed

Question 42 marks

$f(x) = x^3 + ax^2 + 5x + 4$ , where $a$ is a constant. When $f(x)$ is divided by $(x - 3)$ the remainder is $10$ . Which equation correctly captures that information?

$9a + 46 = 10$
Correct. The remainder is $f(3) = 27 + 9a + 15 + 4 = 9a + 46$ , and the question states this equals $10$ — giving $a = -4$ . Both halves of the setup matter: the value substituted, and what it is set equal to.
B $9a + 46 = 0$
This sets the substitution to zero, which is the condition for $(x-3)$ to be a *factor* — a different question. The identity is $f(x) \equiv (x-3)q(x) + r$ with $r = 10$ here, not $0$ ; a real report lists "setting equal to $-3$ or $0$ , rather than $3$ " among the commonest errors on this question type (Jan 2024, Q1).
C $9a - 38 = 10$
This is $f(-3) = -27 + 9a - 15 + 4 = 9a - 38$ . The divisor $(x-3)$ is zero at $x = +3$ ; the sign has been taken from the bracket instead of found by solving it.
D $9a + 46 = -10$
The substitution is right and the remainder's sign has been flipped. A remainder of $10$ means the identity ends $+10$ ; nothing in the division introduces a sign change, and the same report records candidates setting a stated remainder to its negative (Jan 2024, Q1).

Traps tested: Remainder equation set to zero · Root sign taken straight from the bracket · Remainder sign flipped

Question 51 mark

A question gives a cubic with one unknown coefficient and states the remainder on division by a linear expression, then asks for that coefficient. What do the examiners' own records say about the two available routes?

Both earn the same marks, but the remainder theorem is recorded as both the more efficient and the more commonly successful route
Correct on both counts. The mark scheme credits a correct method carried out correctly, so neither route is preferred by the scheme — and the report on a real question of this shape states that "the remainder theorem was the most efficient approach, and this was also the most common," while "candidates who chose to use division gave themselves a more difficult task and tended to be less successful" (Jan 2024, Q1).
BOnly algebraic long division earns full marks, because it shows all the working
There is no such rule. A substitution is working — it is the remainder theorem applied — and the same report describes the remainder theorem as the most common approach on a question that was answered well. Showing the substitution written out, rather than only its value, is what makes the method visible.
CThe two routes give different remainders, so the question must specify which to use
They cannot differ. Both are computing the same $r$ in the same identity $f(x) \equiv (\text{divisor})q(x) + r$ ; the mechanism block derives the substitution route *from* the division, so agreement is structural rather than coincidental.
DNeither is reliable without a calculator check of the division
Calculators are permitted throughout this paper, and are useful for evaluating $f$ at a value — but a bare answer with no method shown is a different risk. The general marking guidance's advice is that a learnt method should be quoted before it is used, since an implied method "may be lost if there is any mistake in the working."

Traps tested: Long division preferred over substitution · Believes both methods give different answers · Calculator substituted for shown method

Question 62 marks

$f(x) = x^3 + x^2 - 2x + 12$ . Which of these is a factor of $f(x)$ ?

$(x + 3)$
Correct. $f(-3) = -27 + 9 + 6 + 12 = 0$ , so by the factor theorem $(x+3)$ is a factor. The search was not a guess: the leading coefficient is $1$ and the constant is $12$ , so any linear factor with integer coefficients must be $(x - b)$ with $b$ dividing $12$ — the candidates are $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$ and nothing else, tested smallest first.
B $(x - 3)$
$f(3) = 27 + 9 - 6 + 12 = 42$ , not zero, so $(x-3)$ is not a factor — it leaves a remainder of $42$ . Both $+3$ and $-3$ are on the candidate list; being on the list means worth testing, not confirmed.
C $(x - 2)$
$f(2) = 8 + 4 - 4 + 12 = 20$ , not zero. Once $(x+3)$ is found, the quotient is $x^2 - 2x + 4$ , whose discriminant is $4 - 16 = -12 < 0$ — so there is no second real linear factor at all, and no value of $x$ other than $-3$ makes $f(x)$ zero.
D $(x + 1)$
$f(-1) = -1 + 1 + 2 + 12 = 14$ , not zero. Testing the smallest candidates first is the right order — they are quickest to evaluate — but a nonzero value settles that candidate and moves the search on rather than ending it.

Traps tested: Root sign taken straight from the bracket · Trial values chosen at random

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Mark scheme: Jan 2020 — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA12.

Algebraic Division, the Factor Theorem and the Remainder Theorem

What spec 2.1 actually asks for — and what the formula booklet does not give you

Where both theorems come from — one identity, evaluated at the value that deletes the quotient

Doing the division itself: long division, or writing the identity down and comparing coefficients

Which values are worth trying — why the constant term tells you where to look, instead of guessing

Given that (x−3)(x - 3)(x−3) is a factor of f(x)=2x3−5x2−4x+3f(x) = 2x^3 - 5x^2 - 4x + 3f(x)=2x3−5x2−4x+3, factorise f(x)f(x)f(x) completely — VERIDIAN-original

Complete the chain — show that (x+2)(x + 2)(x+2) is a factor of f(x)=3x3+2x2−7x+2f(x) = 3x^3 + 2x^2 - 7x + 2f(x)=3x3+2x2−7x+2, and factorise f(x)f(x)f(x) completely

Deep dive

Practice this for real

Given that $(x - 3)$ is a factor of $f(x) = 2x^3 - 5x^2 - 4x + 3$ , factorise $f(x)$ completely — VERIDIAN-original

Complete the chain — show that $(x + 2)$ is a factor of $f(x) = 3x^3 + 2x^2 - 7x + 2$ , and factorise $f(x)$ completely