Simplifying Rational Expressions and Algebraic Division

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.

Why crossing out a matching number is sometimes right and sometimes wrong — before the algebra

In plain terms

Take the fraction 12 over 18. You can shrink it, because both 12 and 18 are built by MULTIPLYING: 12 is 6 times 2, and 18 is 6 times 3. That 6 multiplies the whole top and the whole bottom, so splitting it off and cancelling it is safe — 12 over 18 becomes 2 over 3. Check it on a calculator if you like: 12 divided by 18 is 0.666…, and 2 divided by 3 is 0.666… too, the same number, so nothing was lost by cancelling. Now try a fraction that LOOKS similar: 3 plus 9, all over 3 plus 15 — which is 12 over 18 again, just written a longer way, since 3+9 is 12 and 3+15 is 18. It's tempting to spot the 3 sitting in the top and the 3 sitting in the bottom and cross both out, leaving 9 over 15. But check that against the real value: 9 divided by 15 is 0.6, not 0.666…. Crossing out the 3s changed the actual value of the fraction. The reason is that the 3 in "3 plus 9" isn't multiplying the whole top — the top, 12, is made by ADDING 3 and 9 together, not by multiplying something by 3. Cancelling only ever removes a number that multiplies EVERYTHING on that side. It never removes a number that's just one piece being added into a sum, even when the digits match.

The first case worked because 6 was a common FACTOR — a number multiplying the whole numerator and the whole denominator, so dividing both by it left the value exactly unchanged. The second case failed because 3 was only a common TERM — one piece of a sum, 3+9 and 3+15 — and a term buried inside a sum can't be pulled out and cancelled, because removing it changes what's actually being added up on each side. The rule that carries over into algebra is the identical rule, with letters standing in for numbers instead of the numbers themselves: a fraction written as $k$ times something, over $k$ times something else, can be cancelled down by that $k$ , because $k$ is a genuine common factor of the whole top and the whole bottom. A fraction written as $k$ PLUS something, over $k$ PLUS something else, cannot — the matching $k$ 's are only matching terms, exactly like the two 3's above, and crossing them out is the same mistake with letters instead of digits.

Formally

Cancelling $\frac{Ak}{Bk}$ to $\frac{A}{B}$ is legal for any nonzero $k$ , because it divides the whole numerator and the whole denominator by the same quantity, which never changes a fraction's value — $k$ multiplies EVERYTHING on both sides. There is no equivalent rule for a term that's merely added: $\frac{A+k}{B+k}$ cannot be reduced by cancelling the two $k$ 's, because $k$ does not multiply the whole of $A+k$ or the whole of $B+k$ ; it is one piece of a sum, and a sum does not factor apart from its other pieces just because two of its pieces happen to look alike. Simplifying a rational expression $\frac{P(x)}{Q(x)}$ always proceeds the same way as a result: factorise $P(x)$ and $Q(x)$ fully first, so that whatever they share is exposed as an actual multiplying factor, and only then cancel it — never by hunting for matching symbols wherever they happen to sit in the unfactorised expression.

A rational expression only simplifies when a factor cancels

A is one polynomial divided by another, $\frac{P(x)}{Q(x)}$ with $Q(x)$ not identically zero — spec 1.1's own guidance restricts the denominators tested on this paper to linear or quadratic $Q(x)$ , which is why every example below has one of those two shapes. Simplifying one means factorising the numerator and denominator fully, then cancelling whatever factor is common to both.

That word — factor — is the entire rule, and it's worth being precise about why. Cancelling $\frac{Ak}{Bk}$ to $\frac{A}{B}$ is legal for any nonzero $k$ because it's dividing the whole numerator and the whole denominator by the same quantity, which never changes a fraction's value. The rule works because $k$ multiplies EVERYTHING in the numerator and EVERYTHING in the denominator. There is no equivalent rule for a term that's merely added — $\frac{x+3}{x+5}$ cannot be reduced by cancelling the two $x$ 's, because $x$ does not multiply the whole of $x+3$ or the whole of $x+5$ ; it's one piece of a sum, and a sum doesn't factor apart from its other pieces just because you'd like it to.

Worked example: simplify $\frac{x^2-x-6}{x^2-9}$ . Factorise both: $x^2-x-6=(x-3)(x+2)$ and $x^2-9=(x-3)(x+3)$ . The factor $(x-3)$ is common, so $\frac{x^2-x-6}{x^2-9} = \frac{(x-3)(x+2)}{(x-3)(x+3)} = \frac{x+2}{x+3}$ , for $x \neq 3$ and $x \neq -3$ .

Both exclusions matter, and they come from different places. $x=-3$ still breaks the simplified expression — that restriction is visible. $x=3$ does not: $\frac{x+2}{x+3}$ looks perfectly well-defined there, and gives $\frac{5}{6}$ . But the expression you were actually asked to simplify was $\frac{0}{0}$ at $x=3$ — undefined, not $\frac{5}{6}$ — because that's exactly the value that made the cancelled factor $(x-3)$ zero. Cancelling a factor removes it from the page; it does not undo the restriction it created. State exclusions from the ORIGINAL denominator, before any cancelling, or a genuinely undefined point gets reported as a perfectly ordinary value.

A brief recap: the factor and remainder theorems

This isn't new WMA13 content — it's WMA12 spec item 2.1 ("Simple algebraic division; use of the Factor Theorem and the Remainder Theorem"), assumed prior knowledge for the whole of P3: the spec states outright that "a knowledge of the specifications for P1 and P2, their prerequisites and associated formulae, is assumed and may be tested." If it's rusty, here is the whole of it before this lesson builds on it.

Remainder theorem. When a polynomial $f(x)$ is divided by $(x-a)$ , the remainder equals $f(a)$ . Factor theorem. $(x-a)$ is a factor of $f(x)$ if and only if $f(a)=0$ — the is just the 's zero case, given its own name because it's used so often to hunt for factors before dividing at all.

Worked recap: $f(x) = x^3-2x^2-5x+6$ . Try $x=1$ : $f(1) = 1-2-5+6 = 0$ , so $(x-1)$ is a factor. Dividing confirms it and finds the rest: $f(x) = (x-1)(x^2-x-6) = (x-1)(x-3)(x+2)$ .

A real WMA12 examiner report on almost exactly this task is worth carrying forward into WMA13, because the technique tip transfers even though the spec item doesn't: candidates who used the remainder theorem to find a single unknown value did better than those who reached for to do it — "the remainder theorem was the most efficient approach, and this was also the most common," while candidates using division "gave themselves a more difficult task and tended to be less successful." When a question wants one number — a missing coefficient, a check that a given factor is genuine — substituting is very often faster than dividing. When a question wants the FULL quotient, as every example for the rest of this lesson does, substitution alone can't supply that, and the division has to be done.

Mechanism

Why the remainder theorem is true — and why a quadratic divisor is harder than a linear one

Start from the itself, which is a definition, not a discovery: dividing any polynomial $f(x)$ by any nonzero divisor $D(x)$ produces a UNIQUE quotient $Q(x)$ and remainder $R(x)$ satisfying $f(x) \equiv D(x)Q(x) + R(x)$ , with $\deg R < \deg D$ . The $\equiv$ is doing real work here: it means the equation holds for every value of $x$ , not just some — the two sides are literally the same expression, differently arranged. That licenses substituting ANY value of $x$ into it and getting a true statement, which is the entire mechanism behind everything else in this lesson. Take $D(x) = x-a$ , linear, degree 1. Then $\deg R < 1$ forces $R$ to be a constant — call it $R$ , with no $x$ in it at all — so $f(x) \equiv (x-a)Q(x) + R$ for every $x$ . Substitute the one value that kills the first term: at $x=a$ , $f(a) = (a-a)Q(a) + R = 0 + R = R$ . So $R = f(a)$ . That derivation IS the remainder theorem — not a separate fact to memorise alongside the division algorithm, but that identity evaluated at the single point where it collapses to something trivial. The factor theorem falls out immediately: $(x-a)$ divides $f(x)$ exactly precisely when $R=0$ , which by the line above happens exactly when $f(a)=0$ . Now take $D(x)$ quadratic instead, which is what spec 1.1's own guidance says this paper actually tests. $\deg R < 2$ this time, so $R(x)$ can be linear, $R(x) = mx+n$ — TWO unknowns, not zero. Substituting a root of $D(x)$ still gives one true equation, exactly as before, but one equation can't pin down two unknowns. This is the precise reason a quadratic divisor genuinely needs the division carried out in full (or, equivalently, the coefficients of $Q(x)$ and $R(x)$ found by matching coefficients on both sides of the identity) rather than being shortcut by a single clever substitution the way a linear divisor sometimes can be. It is also why the ONLY thing that makes the leftover fraction reducible afterwards is luck of a specific kind: $R(x)$ happening to share an actual factor with $D(x)$ , which is exactly the "show that" pattern the worked chain below walks through in full.

Worked, in full

Show that $\dfrac{x^3+5x^2+11x+10}{x^2+5x+6} \equiv x + \dfrac{5}{x+3}$ , stating the values of $x$ for which the original expression is undefined

01
Check the degrees first: numerator degree 3, denominator degree 2. The numerator's degree is not less than the denominator's, so this is an improper fraction and must be divided before anything else can be done with it. Divide: the first term of the quotient is $x^3 \div x^2 = x$ .
Earns: M1 — attempts algebraic division of the cubic by the quadratic, with a correct leading term for the quotient (the general marking guidance's own trigger for a method mark on division: the leading term of the dividend divided by the leading term of the divisor).
02
Multiply and subtract: $x(x^2+5x+6) = x^3+5x^2+6x$ , and $(x^3+5x^2+11x+10) - (x^3+5x^2+6x) = 5x+10$ . That result has degree 1, below the divisor's degree 2 — the division algorithm's own stopping condition is met, so the division is finished: quotient $x$ , remainder $5x+10$ .
Earns: A1 — correct quotient $x$ and correct remainder $5x+10$ , both required. Dependent on the M1 above, since an accuracy mark can never stand without its method mark having been earned first.
03
This isn't the form the question asked for yet — the leftover fraction $\frac{5x+10}{x^2+5x+6}$ still has the whole quadratic underneath it, and the target has only $(x+3)$ . Factorise both pieces: $x^2+5x+6=(x+2)(x+3)$ , and $5x+10=5(x+2)$ . The factor $(x+2)$ is common, so $\frac{5x+10}{x^2+5x+6} = \frac{5(x+2)}{(x+2)(x+3)} = \frac{5}{x+3}$ .
Earns: dM1 — factorises the remainder and the divisor and cancels their common factor. Dependent on the division mark: there's nothing to factorise until a remainder actually exists.
04
Combine the two halves: $\dfrac{x^3+5x^2+11x+10}{x^2+5x+6} \equiv x + \dfrac{5}{x+3}$ . That is the required identity, complete.
Earns: A1 — the correct final identity, correct answer only, dependent on the cancel-and-simplify method mark above. This is the exact mark real WMA13 examiner reports document candidates losing — not through wrong mathematics, but by writing the unsimplified leftover fraction and stopping there instead of finishing this line.
05
The ORIGINAL denominator was $(x+2)(x+3)$ , not $(x+3)$ alone — so the expression was never defined at $x=-2$ , even though the final simplified form no longer shows any problem there. State both exclusions: $x \neq -2$ and $x \neq -3$ .
Earns: B1 B1 — both excluded values, independent marks (no method needed: they're read directly off the factorised ORIGINAL denominator, before cancelling). This is the single most commonly dropped step in the whole question type, because the cancellation that happens in stage 3 is precisely what hides the value that needs to be reported here.

Source — Examiner report, Oct 2020

"candidates did not continue to factorise the denominator and cancel (x+3) and hence not proving that Q is 5, they just stated it instead"

Complete it yourself

Complete the chain — express $\dfrac{3x^3+2x^2-15x-2}{x^2+x-6}$ in the form $Ax+B+\dfrac{C}{x+3}$

01
Numerator degree 3, denominator degree 2 — this is improper, so it must be divided first. First term of the quotient: $3x^3 \div x^2 = 3x$ .
02
Multiply and subtract: $3x(x^2+x-6) = 3x^3+3x^2-18x$ , and $(3x^3+2x^2-15x-2) - (3x^3+3x^2-18x) = -x^2+3x-2$ . That result is still degree 2 — the same degree as the divisor — so the division isn't finished: divide again, $-x^2 \div x^2 = -1$ , the next term of the quotient.

Marked, line by line

$f(x) = \dfrac{2x^3+7x^2+8x+3}{x^2+4x+3}$ . (a) Show that $f(x) \equiv 2x-1+\dfrac{C}{x+3}$ , stating the value of the constant $C$ . (4) (b) State the two values of $x$ for which $f(x)$ is not defined. (2) (c) Hence solve the equation $f(x) = 2x+5$ . (2) — VERIDIAN-original question, inspired by the structure of real WMA13 items that pair a division part with a domain part and a 'hence' part built on the simplified form (spec 1.1). Not a reproduction of any past-paper question, and the per-line mark allocations below are modelled on verified mark-scheme conventions rather than copied from a real scheme, which for an original question does not exist.

8 marks available

(a) — 4 marks

“Show that” — the answer is already printed above

Matching the printed result isn’t the same as deriving it — real examiner reports describe scripts that adjust flawed working just to still land on it. Write your own full working below before checking it against the mark scheme.

(b) — 2 marks

101
The original denominator is $x^2+4x+3=(x+1)(x+3)$ , zero at $x=-1$ .
Independent accuracy mark for x = −1. No method required — a correctly read value from the factorised ORIGINAL denominator, unaffected by whatever happened in part (a) afterwards.
B1
102
and zero at $x=-3$ .
Independent accuracy mark for x = −3. Both values are required: the simplified form 2x − 1 + 6/(x+3) only shows the second one as an obvious problem — the first is a removable singularity, present in the original expression and easy to forget precisely because the cancellation in (a) hid it from view.
B1

(c) — 2 marks

201
Using the simplified form from (a): $2x-1+\dfrac{6}{x+3} = 2x+5 \Rightarrow \dfrac{6}{x+3} = 6 \Rightarrow x+3 = 1$ .
Method mark for using the simplified form to reduce the equation to one term in x. This is the direct payoff of finishing part (a) properly — the unsimplified form makes the same equation considerably more work, shown in the common wrong path below.
M1
202
$x=-2$ , and this is admissible since $x=-2$ is not one of the excluded values from part (b).
Accuracy mark, correct answer only. Checking the solution against part (b)'s excluded values isn't optional bookkeeping — a solution equal to an excluded value would have to be rejected, since it would make the original expression undefined rather than solve it.
A1

In your own words

In one sentence: why does cancelling a common factor from a rational expression never remove the domain restriction that factor originally created?

Named traps

cancel-and-simplify-step-abandonedfudged-reverse-fit: Confirmed directly, verbatim, against both the examiner report AND the real mark scheme for the same question. Oct 2020 Q9(a) — divide $\frac{x^4-x^3-10x^2+3x-9}{x^2-x-12}$ — awards the last two of its four marks as "M1: Writes the given expression in the required form using $x^2-x-12=(x-4)(x+3)$ ... A1: Correct answer... Note that Q = 5 is given so it must be shown from correct work, not just stated." Candidates who divided correctly still lost both marks: "candidates did not continue to factorise the denominator and cancel (x+3) and hence not prov[e]... that Q is 5, they just stated it instead" — over 70% scored full marks, but the report names this specific abandonment as the reason the rest did not.
given-value-not-verified-by-substitution: A genuinely different failure from the one above — worth telling apart precisely, because the two are marked differently and fixed differently. Not every division question ends in a fraction to simplify: some GIVE you one of the constants outright and mark it as an independent, "shown" accuracy mark for proving that stated value, not for reaching it — the mark-scheme convention from §4 of the facts bank for "the answer is printed on the paper." Verified verbatim, Jan 2025 Q4(a)(ii) — divide $\frac{4x^3+2x^2+3x+8}{x^2+4} \equiv Ax+B+\dfrac{Cx+D}{x^2+4}$ and "show that $D=0$ " — the real mark scheme awards this as "B1*: Fully shows that D = 0 from clear and correct work... they would need to set up (at least) two correct equations and solve, with appropriate substitutions seen, to show that D = 0." The examiner report confirms candidates who found $A$ , $B$ and $C$ correctly still lost this mark: "did not subsequently establish that D = 0 as they did not show the substitution." There is no factor to cancel here — $x^2+4$ has no real linear factor, so "factorise and cancel" is not the fix. The fix is the same "show that" discipline applied to a different target: write down the actual equation the given value must satisfy, substitute into it, and show it holds — not assert the printed value because it is, in fact, printed.
combined-fraction-not-fully-justified: A separate, independently confirmed pattern across at least three series (Oct 2020, Jan 2024, Jan 2025): candidates combine rational expressions over a common denominator and reach the correct final simplified form, but without showing the intermediate working that justifies it — and lose the mark attached to the justification even though the answer on the page is right. This is the general "show that" rule from the paper's own general marking guidance applied to this specific topic: an answer that happens to be correct isn't the same thing as an answer that's been shown to be correct, and only the second earns a 'show that' mark.
method-not-set-up-before-the-arithmetic: The general marking guidance, verified verbatim from the January 2023 mark scheme and cross-checked against October 2023 and June 2022: "Where a method involves using a formula that has been learnt, the advice given in recent examiners' reports is that the formula should be quoted first... Where the formula is not quoted, the method mark can be gained by implication from correct working with values but may be lost if there is any mistake in the working." Algebraic division has no single formula to quote, but the identity it's built on does: writing $f(x) \equiv D(x)Q(x)+R(x)$ before diving into the subtraction protects the method mark the same way quoting the quadratic formula does — implied method from unlabelled working is real credit, but it is credit that a single early slip can destroy entirely.
exact-form-abandoned-for-a-decimal: Verified verbatim, same general marking guidance: "Examiners' reports have emphasised that where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals." A simplified rational expression or a constant found by cancelling factors is an exact algebraic answer by nature — decimalising it (turning $\frac{5}{x+3}$ into something like " $\approx 1.67$ when $x=0$ ", or rounding a found constant) answers a question that wasn't asked and drops marks a correct exact form would have kept.
calculator-technology-cited-as-the-method: The paper-wide rubric, verified verbatim from a real WMA13 question paper: "In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable." Simplification and division questions are exactly where this lands hardest, because a calculator can often produce the simplified fraction directly. Examiner reports across the paper document candidates losing marks for a correct final answer with no algebraic method shown on a question carrying this instruction — the number being right is not the thing being marked.

Six traps, one habit each — the checklist before the graded questions

Each trap above has a single concrete habit that closes it, and none of the six require new mathematics beyond what's already been worked through above — only doing the step that's easiest to skip once the hard part is finished. Before attempting the MCQs below, run through all six against whatever division or simplification is in front of you.

Finished dividing, and the divisor actually factorises? Don't stop at the leftover fraction. Factorise the remainder and the divisor and cancel whatever's common — the worked chain and marked solution above both show this is a separate, separately-marked step, not an automatic consequence of the division being correct.

Asked to "show that" a stated constant equals a specific value, like $D=0$ ? That's a different mark from the one above, and no amount of factorising reaches it if the divisor has no real factor. Set up the actual equation the given value must satisfy — from matching coefficients or substituting a convenient value into the identity — and show it explicitly holds. The value being printed on the question is exactly why writing it down without that equation earns nothing.

Combining fractions over a common denominator? Show the intermediate line — the common denominator written out, the numerators combined — before stating the simplified result. A correct final fraction with no working between the question and the answer is exactly the pattern examiner reports mark down, whatever the answer itself says.

About to divide or combine? Write the identity first — $f(x) \equiv D(x)Q(x)+R(x)$ , or the common-denominator form — before doing any arithmetic on it. Quoting the structure before using it is what protects the method mark if a later line slips.

Reaching a final value? Leave it exact. A cancelled fraction or a constant found by factorising is already in its exact form; converting it to a decimal answers a question that wasn't asked. And if a calculator could get there directly, show the algebraic method anyway — on a paper that prints "you must show all stages of your working," the number being right is not what's being marked.

Retrieval — with feedback on every choice

Question 1

2 marks

Simplify $\dfrac{2x^2+x-6}{4x^2-9}$ .

Question 2

2 marks

$f(x) = x^3+4x^2-3x-18$ and $f(2)=0$ , so $(x-2)$ is a factor. Dividing gives $f(x) \equiv (x-2)(x^2+ax+b)$ . What are $a$ and $b$ ?

Question 3

3 marks

$\dfrac{x^3+3x^2-8x-30}{x^2-x-6} \equiv x+4+\dfrac{C}{x+2}$ . What is $C$ ?

Question 4

1 mark

$f(x)=x^3-3x+1$ is divided by $x^2+1$ , giving $f(x) \equiv (x^2+1)Q(x)+R(x)$ . Without carrying out the division, what can be said for certain about $R(x)$ ?

Question 5

2 marks

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Show that $\dfrac{2x^3+x^2-11x-10}{x^2-x-6} \equiv 2x+3+\dfrac{C}{x-3}$ , stating the value of $C$ .

A candidate uses a calculator's algebra tools to find the correct simplified fraction directly, and writes only " $C=4$ " with no other working. What is the best description of how this scores?

Reference — not a study method, a lookup

Cancel factors only, never terms: (Ak)/(Bk) = A/B needs k multiplying the WHOLE top and bottom.
Improper (deg num ≥ deg denom): divide first. f(x) ≡ D(x)Q(x) + R(x), deg R < deg D.
Remainder theorem: divide by (x−a), remainder is f(a). Factor theorem: (x−a) a factor ⟺ f(a)=0.
"Show that" [fraction simplifies]: finish the division, THEN factorise and cancel the remainder against the divisor.
"Show that" [a given constant = value, e.g. D=0]: no factor to cancel — set up the equation it must satisfy and substitute to show it holds.
Domain: exclusions come from the ORIGINAL denominator, before cancelling — cancelled factors still count.

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, 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, when follow-through applies, how dependent method marks like dM1 work) rather than transcribed from a real mark scheme, which for an original question does not exist.

Question 12 marks

Simplify $\dfrac{2x^2+x-6}{4x^2-9}$ .

$\dfrac{x+2}{2x+3}$ , for $x \neq \frac{3}{2}$ and $x \neq -\frac{3}{2}$
Correct. $2x^2+x-6=(2x-3)(x+2)$ and $4x^2-9=(2x-3)(2x+3)$ — the second is a difference of two squares, $(2x)^2-3^2$ . The factor $(2x-3)$ cancels, leaving $\frac{x+2}{2x+3}$ , but $x=\frac{3}{2}$ is where that cancelled factor was zero and stays excluded regardless.
B $\dfrac{x+2}{2x+3}$ , valid for all real $x$
The simplification is right and the domain statement is missing. Both $x=\frac{3}{2}$ and $x=-\frac{3}{2}$ made the ORIGINAL denominator zero; only the second still shows a problem in the simplified form.
C $\dfrac{x+2}{2x-3}$
This comes from mis-factorising $4x^2-9$ as a perfect square, $(2x-3)^2$ , instead of a difference of two squares. $(2x-3)^2 = 4x^2-12x+9$ , not $4x^2-9$ — the middle term gives it away. $a^2-b^2$ always factors as $(a-b)(a+b)$ , two DIFFERENT brackets, never one bracket squared.
D $\dfrac{2}{3}$
This treats the leading coefficients (2 and 4) and the constant terms (−6 and −9) as separately cancellable, without ever factorising either quadratic. Rational expressions only simplify through a shared FACTOR of the whole numerator and the whole denominator — comparing individual coefficients like this isn't a valid operation on either one.

Traps tested: Domain exclusions dropped after cancelling · Difference of squares misfactorised as a perfect square · Cancels coefficients and constants separately not factors

Question 22 marks

$f(x) = x^3+4x^2-3x-18$ and $f(2)=0$ , so $(x-2)$ is a factor. Dividing gives $f(x) \equiv (x-2)(x^2+ax+b)$ . What are $a$ and $b$ ?

$a=6$ , $b=9$
Correct. Expand $(x-2)(x^2+ax+b) = x^3+(a-2)x^2+(b-2a)x-2b$ and match coefficients against $x^3+4x^2-3x-18$ : $a-2=4 \Rightarrow a=6$ ; $-2b=-18 \Rightarrow b=9$ ; check the x-term, $b-2a = 9-12=-3$ . ✓ ( $f(x)=(x-2)(x+3)^2$ .)
B $a=4$ , $b=-3$
These are the original cubic's own $x^2$ - and $x$ -coefficients, copied across without actually dividing. The quotient's coefficients are a genuinely different set of numbers from the dividend's — matching coefficients (or dividing directly) is not optional here.
C $a=6$ , $b=-9$
The value of a is right; b has the wrong sign. $-2b=-18$ gives $b=9$ , not $b=-9$ — dividing both sides of $-2b=-18$ by $-2$ flips nothing about the sign of 18 itself, since $-18 \div -2 = +9$ .
D $a=2$ , $b=9$
b is right; a has the wrong sign. $a-2=4$ solves to $a=4+2=6$ , not $a=4-2=2$ — moving the $-2$ to the other side of the equation reverses its sign, and that step was skipped.

Traps tested: Original coefficients copied without dividing · Sign error solving for the constant term · Sign error rearranging the coefficient equation

Question 33 marks

$\dfrac{x^3+3x^2-8x-30}{x^2-x-6} \equiv x+4+\dfrac{C}{x+2}$ . What is $C$ ?

$C=2$
Correct. Dividing gives quotient $x+4$ , remainder $2x-6$ . The divisor factorises as $x^2-x-6=(x-3)(x+2)$ , and the remainder factorises as $2x-6=2(x-3)$ — the factor $(x-3)$ cancels, leaving $\frac{2x-6}{x^2-x-6} = \frac{2}{x+2}$ .
B $C = 2x-6$
This is the remainder BEFORE the cancel-and-simplify step, not after it. $\frac{2x-6}{x^2-x-6}$ still has the whole quadratic underneath — the question's own target form has only $(x+2)$ in the denominator, which means the factorising-and-cancelling step still has to be done.
C $C = -2$
The size is right and the sign has flipped somewhere in the cancellation. $\frac{2(x-3)}{(x-3)(x+2)}$ cancels to $\frac{2}{x+2}$ directly — there is no step in this cancellation that should introduce a sign change, since both the numerator's and denominator's shared factor are identically $(x-3)$ , not $(x-3)$ against $-(x-3)$ .
D $C = 6$
This reads the 6 out of the remainder $2x-6$ directly, ignoring its sign and skipping the factorising step that would show it's $2(x-3)$ , not a standalone constant 6. The remainder has to be factorised before any part of it can be read off as the answer.

Traps tested: Cancel and simplify step abandoned · Sign error when cancelling common factor · Remainder term misread as the simplified constant

Question 41 mark

$f(x)=x^3-3x+1$ is divided by $x^2+1$ , giving $f(x) \equiv (x^2+1)Q(x)+R(x)$ . Without carrying out the division, what can be said for certain about $R(x)$ ?

$R(x)$ has degree 0 or 1 — at most linear, since $\deg R < \deg(x^2+1) = 2$
Correct, and it follows from the division algorithm alone, before either polynomial's actual coefficients are used. (Carrying out the division confirms it: $R(x) = 1-4x$ , genuinely linear.)
B $R(x)=0$ , because $x^2+1$ can't be a factor of a cubic
$x^2+1$ has no real roots, which does rule out a REAL LINEAR factor of that specific shape appearing inside $f(x)$ — but it says nothing about whether $x^2+1$ itself divides $f(x)$ exactly, and nothing at all about the remainder's degree, which is fixed by the division algorithm regardless.
C $R(x)$ has the same degree as $Q(x)$
The two have no required relationship — $Q(x)$ 's degree is fixed by the DIVIDEND's degree (here, $3-2=1$ , so $Q(x)$ is linear), while $R(x)$ 's degree is bounded by the DIVISOR's. They're constrained by different things and can easily differ.
DNothing can be said until the division is actually carried out
One thing can always be said from the structure of division alone: the remainder's degree is bounded by the divisor's. That bound is what lets you recognise a division as finished without needing to know the answer first.

Traps tested: Assumes remainder must be zero · Remainder and quotient degrees conflated · Division algorithm structure not recognised

Question 52 marks

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Show that $\dfrac{2x^3+x^2-11x-10}{x^2-x-6} \equiv 2x+3+\dfrac{C}{x-3}$ , stating the value of $C$ .

A candidate uses a calculator's algebra tools to find the correct simplified fraction directly, and writes only " $C=4$ " with no other working. What is the best description of how this scores?

It's at risk — the correct final value alone, with the instruction on this question, does not reliably earn the method marks
Correct. The instruction printed on the question rules out exactly this: a right answer produced with no algebraic method shown. The mark scheme's method marks are for the division and the cancellation being visible on the page, not for the final value alone, however it was obtained.
BFull marks — a correct final answer always earns full credit, regardless of method
This paper explicitly overrides that assumption on flagged questions. The printed instruction — "solutions relying entirely on calculator technology are not acceptable" — exists precisely because a correct answer with no shown method is not treated as automatically safe here.
CZero marks — using a calculator at all is against the rules on this paper
WMA13 is a calculator-allowed paper throughout; only symbolic-algebra/CAS calculators are banned outright, and only specific parts carry the no-calculator-methods instruction. An ordinary scientific calculator used for arithmetic within a shown method is completely normal on this paper.
DFull marks, provided the calculator used wasn't a graphing calculator
The type of calculator isn't what the printed instruction is policing — the presence or absence of shown algebraic working is. A basic scientific calculator used to skip the method entirely is exactly as unsafe here as a graphing one.

Traps tested: Answer without working · Calculator use entirely disallowed · Calculator type substituted for method shown

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.

Examiner report: Oct 2020 · Q9(a) — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA13.

Simplifying Rational Expressions and Algebraic Division

Why crossing out a matching number is sometimes right and sometimes wrong — before the algebra

A rational expression only simplifies when a factor cancels

A brief recap: the factor and remainder theorems

Why the remainder theorem is true — and why a quadratic divisor is harder than a linear one

Show that x3+5x2+11x+10x2+5x+6≡x+5x+3\dfrac{x^3+5x^2+11x+10}{x^2+5x+6} \equiv x + \dfrac{5}{x+3}x2+5x+6x3+5x2+11x+10​≡x+x+35​, stating the values of xxx for which the original expression is undefined

Complete the chain — express 3x3+2x2−15x−2x2+x−6\dfrac{3x^3+2x^2-15x-2}{x^2+x-6}x2+x−63x3+2x2−15x−2​ in the form Ax+B+Cx+3Ax+B+\dfrac{C}{x+3}Ax+B+x+3C​

Six traps, one habit each — the checklist before the graded questions

Deep dive

Practice this for real

Show that $\dfrac{x^3+5x^2+11x+10}{x^2+5x+6} \equiv x + \dfrac{5}{x+3}$ , stating the values of $x$ for which the original expression is undefined

Complete the chain — express $\dfrac{3x^3+2x^2-15x-2}{x^2+x-6}$ in the form $Ax+B+\dfrac{C}{x+3}$