Sec 3 A-Math Polynomials & Partial Fractions

Overview

What you’ll learn

Divide a polynomial by a linear factor using long division or comparing coefficients, stating the quotient and remainder.
Use the remainder theorem to find remainders without dividing, including with divisors of the form ax − b.
Use the factor theorem to test candidates and factorise cubic polynomials, then solve cubic equations.
Apply the identities for a³ + b³ and a³ − b³.
Express proper fractions with denominators (ax + b)(cx + d), (ax + b)(cx + d)² and (ax + b)(x² + c²) as partial fractions.

In 30 seconds

1Division identity: P(x) = (divisor)(quotient) + remainder, with the remainder of degree lower than the divisor.
2Remainder theorem: the remainder on dividing P(x) by (x − a) is P(a); by (ax − b) it is P(b/a).
3Factor theorem: (x − a) is a factor of P(x) exactly when P(a) = 0 — test easy candidates first: ±1, ±2, ±3.
4Cube identities: a³ + b³ = (a + b)(a² − ab + b²) and a³ − b³ = (a − b)(a² + ab + b²).
5Partial fractions only apply to proper fractions — divide first if the numerator degree is not smaller.
6One constant per factor: A/(linear) for each distinct linear factor, two constants for a repeated factor (one per power).
7A quadratic factor x² + c² that will not factorise over the reals takes a linear numerator: (Ax + B)/(x² + c²).
8After substituting to find constants, compare coefficients to mop up the last one — never guess it.

Notes · 1 of 5

Multiplying out and dividing polynomials

Every division hides the identity P(x) = (divisor)(quotient) + remainder — long division just finds the missing pieces.

A polynomial in x is a sum of terms a xⁿ; its degree is the highest power with a non-zero coefficient. Dividing a polynomial works like long division with numbers, matching the highest powers first.

1Match the leading term
Divide the leading term of the working by the leading term of the divisor; write that above.
2Multiply and subtract
Multiply the whole divisor by your new term, subtract, and bring the degree down.
3Repeat
Stop when the leftover has smaller degree than the divisor — that leftover is the remainder.
4Assemble
P(x) = (divisor)(quotient) + remainder. Check by expanding back.

A faster alternative for linear divisors is comparing coefficients: write P(x) = (x − a)(qx² + rx + s) + R with unknowns, expand the right side, and match coefficients one by one. Both methods earn full marks; choose whichever produces fewer slips for you.

Notes · 2 of 5

The remainder theorem

The remainder of a division by a linear factor is one substitution away.

From the division identity P(x) = (x − a)Q(x) + R, substitute x = a: the huge product (a − a)Q(a) dies, leaving R = P(a). No division needed.

Remainders without dividing
Divisor	Remainder	Example
x − a	P(a)	P(x) = x³ − 2x² + 5x − 1 ÷ (x + 1): P(−1) = −9
x + a	P(−a)	x = −a makes x + a zero
ax − b	P(b/a)	x³ + 2x² − 4 ÷ (2x − 1): P(1/2) = −27/8

Two remainder conditions give two equations — the standard route to unknown coefficients in P(x), used in the harder questions of this chapter.

Notes · 3 of 5

The factor theorem and cracking cubics

P(a) = 0 means (x − a) is a factor — hunt one root, then the quadratic finishes the job.

The factor theorem is the remainder theorem with remainder zero: (x − a) is a factor of P(x) if and only if P(a) = 0. For a cubic with integer coefficients, at least one root is designed to be an easy integer — try ±1, ±2, ±3 and factors of the constant term.

1Test candidates
Substitute 1, −1, 2, −2, … until P equals 0. That value is a root.
2Divide by the factor
Long division or coefficient comparison gives a quadratic quotient.
3Factorise the quadratic
Factorise (or use the formula) to finish.
4State all roots
Solving the cubic means listing every root of the fully factorised form.

Two identities do the same job for perfect cubes of a binomial: a³ + b³ = (a + b)(a² − ab + b²) and a³ − b³ = (a − b)(a² + ab + b²). Spot them by the shapes x³ + 8 = x³ + 2³ and x³ − 27 = x³ − 3³. Careful with the signs: the quadratic inside does not simply drop the cubes — the middle term flips sign relative to the linear factor.

Notes · 4 of 5

Partial fractions: distinct linear factors

Reverse-add a fraction into one piece per factor, with constants found by substitution.

Adding fractions with unlike denominators produces a single fraction; partial fractions undo that. A proper fraction (numerator degree below denominator degree) with denominator (ax + b)(cx + d) splits as A/(ax + b) + B/(cx + d).

1Check the fraction is proper
Numerator degree must be strictly less than the denominator degree — otherwise divide first.
2Write the decomposition
One constant per distinct factor, over each factor.
3Multiply through
Multiply both sides by the full denominator to clear fractions.
4Substitute smart x values
Each factor’s own root kills all but one constant: x = root of (ax + b) finds its constant instantly.
5Compare coefficients
If a constant is still unknown, match x² or constant coefficients.

A quick numerical check against the original fraction (try x = 0 on both sides) confirms the constants before you move on.

Notes · 5 of 5

Repeated factors and quadratic factors

A repeated factor earns one extra constant per extra power; an irreducible quadratic earns a linear numerator.

For a denominator (ax + b)(cx + d)², the repeated factor (cx + d)² needs two pieces: one over (cx + d) and one over (cx + d)². The syllabus’s third shape, (ax + b)(x² + c²), cannot split x² + c² over the reals, so that factor takes a linear numerator (Ax + B).

The three syllabus forms
Denominator	Decomposition form
(ax + b)(cx + d)	A/(ax + b) + B/(cx + d)
(ax + b)(cx + d)²	A/(ax + b) + B/(cx + d) + C/(cx + d)²
(ax + b)(x² + c²)	A/(ax + b) + (Bx + C)/(x² + c²)

Revise

Key terms

The words examiners expect you to use precisely.

Polynomial: A sum of terms a xⁿ with constant coefficients, such as 2x³ − 3x² + x − 6.
Degree: The highest power of x with a non-zero coefficient.
Quotient: The polynomial found by division, before any remainder is added.
Remainder: What is left after division, of degree lower than the divisor.
Remainder theorem: The remainder on dividing P(x) by x − a equals P(a).
Factor theorem: (x − a) is a factor of P(x) exactly when P(a) = 0.

Show 5 more key terms

Root: A value of x for which the polynomial equals zero.
Partial fractions: Writing one fraction as a sum of simpler fractions, one per factor of the denominator.
Proper fraction: A fraction whose numerator has smaller degree than its denominator.
Repeated factor: A factor appearing with a power, such as (cx + d)²; it earns one constant per power.
Irreducible quadratic: A quadratic such as x² + c² with no real linear factors; it takes a linear numerator in partial fractions.

Revise

Common mistakes

Where students most often lose marks in this chapter — and what to do instead.

Mistake: Dividing without keeping columns aligned, e.g. treating x³ + 2x² − 5 as x³ + 2x² + 5.
The missing x term must be written as 0x so subtractions line up.
Instead: Insert placeholders for every missing power before long division.
Mistake: Using P(a) instead of P(−a) for a divisor x + a.
The divisor vanishes at x = −a, so that substitution is what kills the quotient.
Instead: Rewrite x + a as x − (−a): the remainder is P(−a).

Show 8 more mistakes

Mistake: Substituting b instead of b/a for a divisor ax − b.
ax − b = 0 at x = b/a; substituting b evaluates somewhere meaningless.
Instead: Solve divisor = 0 first, then substitute that root.
Mistake: Stopping after one root and calling the quadratic quotient "solved".
A cubic equation has up to three roots; the quotient still factorises.
Instead: Divide by the found factor, factorise the quadratic, list all roots.
Mistake: Writing a³ − b³ = (a − b)(a² − ab + b²) with the wrong middle sign.
The middle term of the quadratic has the opposite sign to the linear factor: (a − b)(a² + ab + b²).
Instead: Expand your factorisation as a check — the cross terms must cancel the cubes to leave only a³ − b³.
Mistake: Applying partial fractions to an improper fraction like (2x³ − 3x + 4)/(x − 1) directly.
The numerator degree is not smaller, so the A/(x − 1) form misses the polynomial part.
Instead: Divide first: the answer is quotient + remainder/divisor, here 2x² + 2x − 1 + 3/(x − 1).
Mistake: Giving a repeated factor only one constant: A/((x + 1)²) with no A/(x + 1) companion.
Two different fractions — 1/(x + 1) and 1/(x + 1)² — cannot merge, so both pieces are needed.
Instead: One constant per power: B/(x + 1) + C/(x + 1)².
Mistake: Giving a quadratic factor a constant numerator: A/(x² + 1) instead of (Ax + B)/(x² + 1).
A constant over a quadratic cannot rebuild every numerator the identity needs.
Instead: Irreducible quadratic factors take a linear numerator (Ax + B).
Mistake: Finding A and C by substitution, then guessing B.
Guessing is not a method and fails when the numbers are less friendly.
Instead: Compare coefficients (usually x²) — the equation left after substitutions pins B exactly.
Mistake: Multiplying only one factor when clearing denominators in the identity.
The identity holds for all x; it must be multiplied by the whole denominator.
Instead: Multiply through by (x − a)(x + b)… in full, then substitute the roots one at a time.

Check yourself

Quick quiz

18 questions with instant explanations. About 5 minutes.

Question 1 of 3

Question 1: current question
Question 2: not answered
Question 3: not answered

Identify polynomial degree

What is the degree of the polynomial 4x³ − x⁵ + 2?

Practise

Topical questions

Exam-style practice in three tiers, from recall to data questions.

5Foundation
5Standard
5Challenging

15 questions worth 67 marks, each with a model answer, mark scheme and examiner tips.

Practise topical questions

Exam practice

Practice paper

A 30-mark paper set out like a school paper, with a separate mark scheme. Give yourself 40 min.

Question paper

Chapter 3 Practice Paper

30 marks · 40 min · 5 pages

5 multiple-choice questions and structured questions with answer lines. Chapter 3 (A4 Polynomials and partial fractions).

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Mark scheme

PRO

Chapter 3 Practice Paper — mark scheme

Answers · Marking points · 3 pages

Model answers with a mark for each creditworthy point — for marking at home or in class.

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Where this chapter is tested

Each school sets its own scope — check your assessment outline.

Questions

Frequently asked questions

What does Sec 3 A-Math Chapter 3 cover?

It covers syllabus sub-topic A4 of O-Level 4049 / SEC G3 K341: polynomial multiplication and division, the remainder and factor theorems, factorising and solving cubics, the sum and difference of cubes, and partial fractions of the three syllabus forms — distinct linear factors, a repeated linear factor, and a linear factor with an irreducible quadratic.

How do I start solving a cubic equation?

Test small integer candidates (±1, ±2, ±3, and factors of the constant term) in the polynomial. When one gives zero, the factor theorem makes it a factor; divide it out and factorise the remaining quadratic.

When do I need to divide before doing partial fractions?

Whenever the numerator’s degree is equal to or greater than the denominator’s. Divide first, write polynomial + remainder/denominator, and only the remainder part is decomposed further.

Why does a repeated factor need two constants?

Fractions like 1/(x + 1) and 1/(x + 1)² are genuinely different and cannot combine into one term, so the decomposition needs one piece per power of the repeated factor.

When is this chapter usually tested in school?

It typically appears in Weighted Assessments 1 and 2 and again in the end-of-year examination. Scope and weighting vary by school, so check your school’s assessment outline.

Sec 3 A-Math Polynomials & Partial Fractions notes