Grade 12 · Pure Mathematics 2 · Cambridge A-Level 9709 · Age 17–18
Welcome to A-Level Algebra
Polynomial algebra sits at the heart of Pure Mathematics 2. Mastering polynomial division, the remainder and factor theorems, and partial fractions unlocks powerful tools used throughout A-Level: from integrating rational functions in Pure Mathematics to modelling in applied mathematics. Every topic in this module feeds directly into calculus and further pure work.
Division Algorithm: f(x) = d(x) · q(x) + r(x) where deg(r) < deg(d), or r is a constant
In this module you will learn to divide polynomials systematically, determine remainders without performing full division, identify factors, and decompose rational expressions into simpler partial fractions ready for integration.
Learning Objectives
Perform polynomial long division and state quotient and remainder
Apply synthetic division as an efficient alternative method
State and apply the Remainder Theorem: remainder = f(a) when dividing by (x − a)
State and apply the Factor Theorem: (x − a) is a factor iff f(a) = 0
Test candidate factors systematically to fully factorise a polynomial
Decompose rational expressions with distinct linear factors into partial fractions
Handle repeated linear factors in partial fraction decomposition
Handle irreducible quadratic factors of the form ax² + bx + c
Recognise and handle improper fractions (divide first before decomposing)
Apply partial fractions to integration and other A-Level contexts
Topics in This Module
Polynomial Division
Long division and synthetic division algorithms
Remainder Theorem
Finding remainders without full division
Factor Theorem
Testing and finding factors of polynomials
Partial Fractions
Distinct and repeated linear denominators
Quadratic Denominators
Irreducible quadratic factors in PF
Improper Fractions
Dividing first, then decomposing
Learn 1 — Polynomial Division
When we divide a polynomial f(x) by a divisor d(x), we obtain a quotient q(x) and remainder r(x) such that:
f(x) = d(x) · q(x) + r(x)
The degree of r(x) is strictly less than the degree of d(x). When dividing by a linear factor (x − a), the remainder is a constant.
Long Division Algorithm
To divide 2x³ + 3x² − x + 4 by (x + 2):
Step 1: Divide the leading term of the dividend by the leading term of the divisor: 2x³ ÷ x = 2x². Write 2x² above the line.
Step 2: Multiply the entire divisor by 2x²: 2x²(x + 2) = 2x³ + 4x². Subtract from the dividend: (2x³ + 3x² − x + 4) − (2x³ + 4x²) = −x² − x + 4
Step 3: Bring down and repeat. Divide −x² by x = −x. −x(x + 2) = −x² − 2x. Subtract: (−x² − x + 4) − (−x² − 2x) = x + 4
Step 4: Divide x by x = 1. 1·(x + 2) = x + 2. Subtract: (x + 4) − (x + 2) = 2
Result: 2x³ + 3x² − x + 4 = (x + 2)(2x² − x + 1) + 2 Quotient: 2x² − x + 1, Remainder: 2
Watch out: If any degree is missing in the dividend (e.g., no x² term), insert a placeholder 0x² to keep alignment correct.
Synthetic Division
Synthetic division is a shortcut when dividing by a linear factor (x − k). Write only the coefficients and work with the value k directly.
To divide 2x³ + 3x² − x + 4 by (x + 2), use k = −2:
Tip: Synthetic division only works when dividing by a linear factor (x − k). For quadratic divisors, use long division.
Learn 2 — Remainder Theorem
Remainder Theorem: If a polynomial f(x) is divided by (x − a), the remainder is f(a).
Proof
By the division algorithm: f(x) = (x − a) · q(x) + r, where r is a constant.
Substitute x = a: f(a) = (a − a) · q(a) + r = 0 + r = r. Hence remainder = f(a). ∎
Example 1 — Finding a remainder directly
Find the remainder when f(x) = x³ − 4x + 7 is divided by (x − 3).
Remainder = f(3) = 27 − 12 + 7 = 22
f(x) = 2x³ + ax² + bx − 1. Given f(1) = 4 and f(−1) = −6, find a and b.
f(1): 2 + a + b − 1 = 4 → a + b = 3 … (i)
f(−1): −2 + a − b − 1 = −6 → a − b = −3 … (ii)
(i) + (ii): 2a = 0 → a = 0; then b = 3
Extended Remainder Theorem
When dividing by (ax − b), the remainder is f(b/a).
Example: remainder when dividing by (2x − 1) is f(1/2).
Tip: Always substitute x = a (not x = −a) when dividing by (x − a). If dividing by (x + 2), substitute x = −2.
Learn 3 — Factor Theorem
Factor Theorem: (x − a) is a factor of f(x) if and only if f(a) = 0.
Proof
(⇒) If (x − a) is a factor, then f(x) = (x − a)q(x), so f(a) = 0 · q(a) = 0.
(⇐) If f(a) = 0, then by the Remainder Theorem the remainder when dividing by (x − a) is 0, so (x − a) divides f(x) exactly, making it a factor. ∎
Strategy: Fully Factorise a Cubic
Step 1: Try integer values ±1, ±2, ±3, … that are factors of the constant term. Find a value a where f(a) = 0.
Step 2: Divide f(x) by (x − a) using long division or synthetic division to get a quadratic quotient q(x).
Step 3: Factorise q(x) using the quadratic formula or inspection.
To solve x³ − 6x² + 11x − 6 = 0:
From the factorisation: (x − 1)(x − 2)(x − 3) = 0
Solutions: x = 1, x = 2, x = 3
Tip: For f(x) = ax³ + … + d, always try factors of d divided by factors of a as potential rational roots.
Learn 4 — Partial Fractions (Linear Denominators)
Partial fractions allow us to write a complicated rational expression as a sum of simpler ones. This is especially useful for integration. The key rule: the degree of the numerator must be strictly less than the degree of the denominator (proper fraction).
Case 1: Distinct Linear Factors
f(x) / [(x+a)(x+b)] = A/(x+a) + B/(x+b)
Cover-up method: To find A, cover (x+a) in the original fraction and substitute x = −a. For B, cover (x+b) and substitute x = −b.
Example: Express (5x + 1) / [(x+1)(x+2)] in partial fractions.
A: cover (x+1), set x = −1: (−5+1)/(−1+2) = −4/1 = −4 → A = −4
B: cover (x+2), set x = −2: (−10+1)/(−2+1) = −9/−1 = 9 → B = 9 Answer: −4/(x+1) + 9/(x+2)
Equating Coefficients Method
Write (5x+1) = A(x+2) + B(x+1)
Expand: (A+B)x + (2A+B) = 5x + 1
Coefficients of x: A + B = 5
Constants: 2A + B = 1
Solving: A = −4, B = 9 ✓
For a repeated factor (x+a)², you need both A/(x+a) and B/(x+a)² terms. Use the cover-up method for the highest power factor, then equate coefficients or substitute another value for the remaining constants.
Example: Express (3x + 1) / [(x−1)²(x+1)] in partial fractions.
3x + 1 = A(x−1)(x+1) + B(x+1) + C(x−1)²
Set x = 1: 4 = 2B → B = 2
Set x = −1: −2 = 4C → C = −1/2
Equate x² coefficients: 0 = A + C → A = 1/2 Answer: (1/2)/(x−1) + 2/(x−1)² − (1/2)/(x+1)
Common error: For a repeated factor (x+a)², you must include BOTH A/(x+a) and B/(x+a)². Omitting the first term leads to an incorrect result.
When the denominator contains an irreducible quadratic factor (one that cannot be factorised over the reals, i.e. discriminant < 0), the numerator for that term must be linear: (Ax + B).
Example: Express (2x² + x + 1) / [(x²+1)(x+1)] in partial fractions.
Write: (2x²+x+1) = (Ax+B)(x+1) + C(x²+1)
Set x = −1: 2 − 1 + 1 = 2C → C = 1
Expand: (Ax+B)(x+1) + (x²+1) = 2x²+x+1
(A+1)x² + (A+B)x + (B+1) = 2x²+x+1
A+1 = 2 → A = 1; B+1 = 1 → B = 0; check: A+B = 1 ✓ Answer: x/(x²+1) + 1/(x+1)
Improper Fractions
A fraction is improper when deg(numerator) ≥ deg(denominator). You must perform polynomial division first to extract the polynomial part, then decompose the proper remainder.
If deg(f) ≥ deg(d): first write f(x)/d(x) = quotient + remainder/d(x), then decompose remainder/d(x)
Example: Express (x³ + 2x − 3) / (x² − 1) as partial fractions.
Degree of numerator (3) ≥ degree of denominator (2) → improper.
Divide: x³ + 2x − 3 = x(x²−1) + (3x − 3)
So: (x³+2x−3)/(x²−1) = x + (3x−3)/(x²−1)
Now decompose (3x−3)/[(x−1)(x+1)]: A/(x−1) + B/(x+1)
x = 1: 0 = 2A → A = 0
x = −1: −6 = −2B → B = 3 Answer: x + 0/(x−1) + 3/(x+1) = x + 3/(x+1)
Check: Always verify your partial fractions are correct by recombining them over a common denominator and confirming you recover the original expression.
Fully Worked Examples
Example 1 — Long Division M1 A1 A1
Divide 3x³ − 2x² + 5x − 1 by (x − 2). Find quotient and remainder.
M1: Set up long division with 3x³ − 2x² + 5x − 1 ÷ (x − 2).
3x³ ÷ x = 3x². Multiply: 3x²(x−2) = 3x³ − 6x². Subtract: 4x² + 5x − 1.
A1: Proper part: (3x−3)/[(x−1)(x+1)] = A/(x−1) + B/(x+1)
x = 1: 0 = 2A → A = 0
x = −1: −6 = −2B → B = 3
M1 A1:Final answer: x + 3/(x+1)
Common Mistakes
1. Missing placeholder for absent terms
Wrong: Dividing x³ + 2x + 1 by (x−1) without noticing there is no x² term — misaligning the division.
Right: Write x³ + 0x² + 2x + 1. Insert the 0x² placeholder before starting long division to keep columns aligned.
2. Wrong substitution value in remainder theorem
Wrong: When dividing by (x + 2), using x = 2 in f(x) to find the remainder.
Right: (x + 2) = (x − (−2)), so substitute x = −2. Remainder = f(−2).
3. Confusing remainder theorem and factor theorem
Wrong: Saying "(x−a) is a factor because f(a) = 5" — any non-zero remainder means it is NOT a factor.
Right: (x−a) is a factor if and only if f(a) = 0. The factor theorem is the special case of the remainder theorem where r = 0.
4. Incomplete form for repeated factors
Wrong: Writing f(x)/[(x+a)²(x+b)] = A/(x+a)² + B/(x+b) — the A/(x+a) term is missing.
Right: Must use A/(x+a) + B/(x+a)² + C/(x+b). Each power of the repeated factor up to and including the highest power needs its own term.
5. Forgetting to divide first for improper fractions
Wrong: Attempting to write (x²+3x+1)/(x²−1) = A/(x−1) + B/(x+1) directly — this gives inconsistent equations because the degree of numerator equals the degree of denominator.
Right: Since deg(numerator) ≥ deg(denominator), first divide to get a polynomial plus a proper fraction, then decompose the proper remainder.
6. Cover-up method: substituting the wrong value
Wrong: For A/(x+3), using x = 3 instead of x = −3 in the cover-up method.
Right: Set the factor equal to zero: x + 3 = 0 → x = −3. Always solve the factor for x, then substitute that x value into the remaining expression.
7. Not verifying partial fractions by recombining
Wrong: Stopping once you have A, B, C values without checking if they actually reproduce the original fraction.
Right: Recombine your partial fractions over a common denominator and verify you get the original numerator. Takes 30 seconds and catches arithmetic errors.
If deg(numerator) ≥ deg(denominator): divide first, then decompose
Improper Condition
Fraction is improper when deg(f) ≥ deg(d)
Proof Bank
Proof 1 — Remainder Theorem
Theorem: If f(x) is a polynomial and is divided by (x − a), the remainder is f(a).
Proof: By the division algorithm, there exist a polynomial q(x) and a constant r such that:
f(x) = (x − a) · q(x) + r
This identity holds for all values of x. Substituting x = a:
f(a) = (a − a) · q(a) + r = 0 · q(a) + r = r
Therefore the remainder r = f(a). ∎
Proof 2 — Factor Theorem (both directions)
Theorem: (x − a) is a factor of f(x) if and only if f(a) = 0.
Proof (⇒): Suppose (x − a) is a factor of f(x). Then there exists a polynomial q(x) such that f(x) = (x − a)q(x). Substituting x = a: f(a) = (a − a)q(a) = 0. ✓
Proof (⇐): Suppose f(a) = 0. By the Remainder Theorem, when f(x) is divided by (x − a), the remainder is f(a) = 0. By the division algorithm: f(x) = (x − a)q(x) + 0 = (x − a)q(x). Therefore (x − a) is a factor of f(x). ✓
Both directions are proven, so the biconditional holds. ∎
Proof 3 — Uniqueness of Partial Fraction Decomposition
Claim: The partial fraction decomposition of a proper rational function with distinct factors is unique.
Proof (sketch): Suppose p(x)/[(x+a)(x+b)] has two decompositions:
A/(x+a) + B/(x+b) = A'/(x+a) + B'/(x+b)
Multiply through by (x+a)(x+b):
A(x+b) + B(x+a) = A'(x+b) + B'(x+a)
This is a polynomial identity. Comparing coefficients of x: A + B = A' + B', and constants: Ab + Ba = A'b + B'a.
Setting x = −a: A(−a+b) = A'(−a+b). Since a ≠ b, we divide: A = A'. Similarly B = B'. Hence the decomposition is unique. ∎
Polynomial Evaluator & Remainder Visualiser
Enter the coefficients of f(x) = ax³ + bx² + cx + d and a value k. The visualiser computes f(k) using the Remainder Theorem and tells you whether (x − k) is a factor.
Enter values above and click Evaluate f(k).
Division Result
Partial Fraction Checker (Distinct Linear)
For a fraction (px + q) / [(x+a)(x+b)], enter the values to compute A and B using the cover-up method.