Algebraic methods at A-Level
This topic gathers four skills that recur constantly through the rest of A-Level: manipulating algebraic fractions using your GCSE factorising skills, dividing polynomials, using the factor theorem to factorise cubics, and writing formal mathematical proofs — a step up from the informal justification expected at GCSE.
Four skills make up this topic:
- Algebraic fractions — simplifying by factorising, and adding/subtracting using a common denominator
- Polynomial division — dividing a cubic (or higher) by a linear factor, finding the quotient and remainder
- The factor theorem — using \(f(a) = 0 \iff (x-a)\) is a factor, to fully factorise a cubic
- Mathematical proof — proof by deduction, proof by exhaustion, and disproof by counter-example
Algebraic fractions
To simplify an algebraic fraction, factorise the numerator and denominator fully, then cancel any factor common to both. To add or subtract algebraic fractions, find a common denominator exactly as you would with numerical fractions.
Worked Example 1
Simplify fully: \(\dfrac{x^2 - 9}{x^2 + x - 6}\)
1
Factorise the numerator: \(x^2 - 9 = (x-3)(x+3)\)
2
Factorise the denominator: \(x^2+x-6 = (x+3)(x-2)\)
3
Cancel the common factor \((x+3)\)
Answer\(\dfrac{x-3}{x-2}\)
Worked Example 2
Express as a single fraction: \(\dfrac{2}{x+1} + \dfrac{3}{x-2}\)
1
Use the common denominator \((x+1)(x-2)\)
2
\(\dfrac{2(x-2) + 3(x+1)}{(x+1)(x-2)}\)
3
Expand the numerator: \(2x-4+3x+3 = 5x-1\)
Answer\(\dfrac{5x-1}{(x+1)(x-2)}\)
Polynomial division
Dividing a polynomial by a linear expression \((x - a)\) works exactly like long division of numbers: divide, multiply, subtract, bring down — repeated until nothing is left to bring down. The result is a quotient and, if it doesn't divide exactly, a remainder.
Worked Example 3
Divide \(2x^3 - 3x^2 - 11x + 6\) by \((x-3)\), giving the quotient.
1
Divide the leading terms to find the first quotient term, \(2x^2\) — subtracting its product with the divisor clears the \(x^3\) term and reveals the next one.
2
Repeat on the new leading term: dividing gives the next quotient term, \(+3x\), which clears it in the same way.
3
One more cycle finds the last quotient term, \(-2\), and clears the rest exactly — a remainder of \(0\) confirms \((x-3)\) is a genuine factor of the cubic.
\(x - 3\)
\(2x^2\)
\(+3x\)
\(-2\)
\(2x^3\)
\(-3x^2\)
\(-11x\)
\(+6\)
\(2x^3\)
\(-6x^2\)
\(3x^2\)
\(-11x\)
\(3x^2\)
\(-9x\)
\(-2x\)
\(+6\)
\(-2x\)
\(+6\)
\(0\)
Full working for the division above, building the quotient \(2x^2+3x-2\) one term at a time.
AnswerQuotient \(= 2x^2+3x-2\), remainder \(0\)
The factor theorem
The factor theorem states: \(f(a) = 0\) if and only if \((x-a)\) is a factor of \(f(x)\). This turns "fully factorise this cubic" into a three-stage process: test values of \(a\) to find one root, divide to reduce the cubic to a quadratic, then factorise the quadratic as usual.
Worked Example 4
Using the factor theorem, fully factorise \(f(x) = 2x^3 - 3x^2 - 11x + 6\).
1
Test \(x=3\): \(f(3) = 54 - 27 - 33 + 6 = 0\), so \((x-3)\) is a factor
2
Dividing (Worked Example 3) gives \(f(x) = (x-3)(2x^2+3x-2)\)
3
Factorise the quadratic: \(2x^2+3x-2 = (2x-1)(x+2)\)
Answer\(f(x) = (x-3)(2x-1)(x+2)\)
Mathematical proof
A-Level proof is more formal than GCSE reasoning, and comes in three types you must recognise:
- Proof by deduction: start from a general algebraic expression (e.g. \(2n\) for "any even number") and show the required result follows logically, step by step.
- Proof by exhaustion: when only a small, finite number of cases are possible, check every single one.
- Disproof by counter-example: to show a statement is false, it's enough to find just one case where it fails — a full proof isn't needed to disprove.
Worked Example 5
Prove that the sum of two consecutive odd numbers is always even.
1
Let the two consecutive odd numbers be \(2n+1\) and \(2n+3\), for integer \(n\)
2
Sum \(= (2n+1)+(2n+3) = 4n+4\)
3
\(4n+4 = 2(2n+2)\), which is \(2 \times\) an integer
AnswerThe sum is always a multiple of 2, so always even, as required
Worked Example 6
Show that the statement "\(n^2+n+1\) is always a prime number, for integer \(n \geq 0\)" is false.
1
A single counter-example is enough to disprove the statement
2
Try \(n=4\): \(4^2+4+1 = 21\)
3
\(21 = 3 \times 7\), which is not prime
Answer\(n=4\) gives 21, which is not prime — the statement is false
Exam tips
💡 Tip 1
Always factorise before cancelling
You can only cancel whole factors, never individual terms — \(\dfrac{x+3}{x}\) does not simplify to \(3\). Factorise numerator and denominator fully first, then cancel.
💡 Tip 2
Test small integer values first for the factor theorem
Try \(\pm1, \pm2, \pm3\) (factors of the constant term) before anything else — the correct root is almost always among these in exam questions.
💡 Tip 3
Check your division by expanding back
After dividing, multiply your quotient by the divisor and add the remainder — it should reconstruct the original polynomial exactly. This catches arithmetic slips before you lose marks.
💡 Tip 4
Start every deduction proof with "Let..."
Define your variables explicitly (e.g. "let the integer be \(n\)") before manipulating anything — mark schemes award a mark for correctly setting up the general case, not just the final algebra.
💡 Tip 5
One counter-example is always enough to disprove
Don't try to "prove it's sometimes false" with general algebra — find a single specific value that breaks the statement and state the result. That alone is a complete disproof.
Common mistakes
Common Mistake 1
Cancelling terms instead of factors
In \(\dfrac{x^2+2x}{x}\), it's tempting to cancel the \(x\) in "\(2x\)" against the denominator — but you must factorise first: \(\dfrac{x(x+2)}{x} = x+2\). Cancelling only works on whole common factors.
Common Mistake 2
Forgetting to expand brackets when adding fractions
In \(\dfrac{2(x-2)+3(x+1)}{(x+1)(x-2)}\), leaving the numerator unexpanded or expanding it incorrectly (e.g. missing a sign) is the single most common lost mark in this topic.
Common Mistake 3
Sign errors during polynomial division
Subtracting a negative term (e.g. \(-2x - (-9x)\)) is where most division mistakes happen — write out the subtraction explicitly rather than doing it mentally.
Common Mistake 4
Confusing "prove" with "verify for one case"
Showing a statement is true for \(n=1\) proves nothing about all \(n\) — a deduction proof needs the general case; only a finite, exhaustively-checked set of cases justifies proof by exhaustion.
Common Mistake 5
Trying to "prove" a false statement instead of disproving it
If asked to show a statement is false, don't attempt algebra to "prove" it — find one counter-example. Attempting a general proof of a false statement wastes time and cannot succeed.
Practice questions
Work through each question before checking the worked solution.
Core Skills
Q1Simplify fully: \(\dfrac{x^2-4}{x^2+5x+6}\)Core Skills
Q2Express as a single fraction: \(\dfrac{3}{x+2} + \dfrac{1}{x-1}\)Core Skills
Q3Divide \(x^3+2x^2-5x-6\) by \((x+3)\), giving the quotient.Core Skills
Q4Show that \((x-2)\) is a factor of \(f(x) = x^3-3x^2-4x+12\).Core Skills
Q5Prove that the sum of two even numbers is always even.Core Skills
Exam-Style
Q6Simplify fully: \(\dfrac{2x^2+x-3}{x^2-1}\)Exam-Style
Q7Express as a single fraction: \(\dfrac{4}{x-3} - \dfrac{2}{x+1}\)Exam-Style
Q8Fully factorise \(f(x) = x^3-2x^2-5x+6\) using the factor theorem.Exam-Style
Q9Divide \(x^3-4x^2+x+7\) by \((x-2)\), stating the quotient and remainder.Exam-Style
Q10Prove that the sum of three consecutive integers is always a multiple of 3.Exam-Style
A* Challenge
Q11Given that \((x+1)\) and \((x-2)\) are both factors of \(f(x) = x^3+ax^2+bx-6\), find the values of \(a\) and \(b\).A* Challenge
Q12Simplify fully: \(\dfrac{x^3-8}{x^2-4}\)A* Challenge
Q13Prove by exhaustion that \(n^2+n\) is even for every integer \(n\) with \(0 \leq n \leq 4\).A* Challenge
Q14Disprove the statement: "for all positive integers \(n\), \(2^n+1\) is prime."A* Challenge
Q15Given that \((x-2)\) is a factor of \(f(x) = 2x^3+x^2-13x+6\), express \(f(x)\) as a product of three linear factors.A* Challenge
Answers — full worked solutions
Core Skills (Q1–Q5)
Q1 — \(\dfrac{x-2}{x+3}\)
1
\(x^2-4 = (x-2)(x+2)\); \(x^2+5x+6 = (x+2)(x+3)\)
2
Cancel the common factor \((x+2)\)
\(\dfrac{x-2}{x+3}\)
Q2 — \(\dfrac{4x-1}{(x+2)(x-1)}\)
1
\(\dfrac{3(x-1) + 1(x+2)}{(x+2)(x-1)}\)
2
\(3x-3+x+2 = 4x-1\)
\(\dfrac{4x-1}{(x+2)(x-1)}\)
Q3 — Quotient \(= x^2-x-2\)
1
\(x^3 \div x = x^2\); \(x^2(x+3)=x^3+3x^2\); subtract to leave \(-x^2-5x\)
2
\(-x^2 \div x = -x\); \(-x(x+3)=-x^2-3x\); subtract to leave \(-2x-6\)
3
\(-2x \div x = -2\); \(-2(x+3)=-2x-6\); subtract to leave remainder \(0\)
Quotient \(= x^2-x-2\), remainder \(0\)
Q4 — Confirmed, \(f(2)=0\)
1
\(f(2) = 8 - 12 - 8 + 12\)
\(f(2) = 0\), so \((x-2)\) is a factor
Q5 — Sum \(= 2(n+m)\), always even
1
Let the two even numbers be \(2n\) and \(2m\), for integers \(n, m\)
2
Sum \(= 2n+2m = 2(n+m)\)
A multiple of 2, so always even, as required
Exam-Style (Q6–Q10)
Q6 — \(\dfrac{2x+3}{x+1}\)
1
\(2x^2+x-3 = (2x+3)(x-1)\); \(x^2-1 = (x-1)(x+1)\)
2
Cancel the common factor \((x-1)\)
\(\dfrac{2x+3}{x+1}\)
Q7 — \(\dfrac{2x+10}{(x-3)(x+1)}\)
1
\(\dfrac{4(x+1) - 2(x-3)}{(x-3)(x+1)}\)
2
\(4x+4-2x+6 = 2x+10\)
\(\dfrac{2x+10}{(x-3)(x+1)}\)
Q8 — \(f(x) = (x-1)(x-3)(x+2)\)
1
\(f(1) = 1-2-5+6 = 0\), so \((x-1)\) is a factor
2
Dividing gives \(f(x) = (x-1)(x^2-x-6)\)
3
\(x^2-x-6 = (x-3)(x+2)\)
\(f(x) = (x-1)(x-3)(x+2)\)
Q9 — Quotient \(= x^2-2x-3\), remainder \(1\)
1
\(x^3 \div x = x^2\); \(x^2(x-2)=x^3-2x^2\); subtract to leave \(-2x^2+x\)
2
\(-2x^2 \div x = -2x\); \(-2x(x-2)=-2x^2+4x\); subtract to leave \(-3x+7\)
3
\(-3x \div x = -3\); \(-3(x-2)=-3x+6\); subtract to leave remainder \(1\)
Quotient \(= x^2-2x-3\), remainder \(1\)
Q10 — Sum \(= 3(n+1)\), always a multiple of 3
1
Let the three consecutive integers be \(n\), \(n+1\), \(n+2\)
2
Sum \(= n+(n+1)+(n+2) = 3n+3\)
3
\(3n+3 = 3(n+1)\)
A multiple of 3, for every integer \(n\), as required
A* Challenge (Q11–Q15)
Q11 — \(a=2\), \(b=-5\)
1
\(f(-1)=0\): \(-1+a-b-6=0 \implies a-b=7\)
2
\(f(2)=0\): \(8+4a+2b-6=0 \implies 2a+b=-1\)
3
Adding \(a-b=7\) and \(2a+b=-1\) (after substitution): \(3a=6 \implies a=2\), then \(b=-5\)
\(a=2\), \(b=-5\)
Q12 — \(\dfrac{x^2+2x+4}{x+2}\)
1
\(x^3-8 = (x-2)(x^2+2x+4)\) (difference of cubes)
2
\(x^2-4 = (x-2)(x+2)\)
3
Cancel the common factor \((x-2)\)
\(\dfrac{x^2+2x+4}{x+2}\)
Q13 — Confirmed even for \(n=0,1,2,3,4\)
1
\(n=0: 0\); \(n=1: 2\); \(n=2: 6\); \(n=3: 12\); \(n=4: 20\)
Every value is even, so proven by exhaustion for \(0 \leq n \leq 4\)
Q14 — \(n=3\) gives \(2^3+1=9=3\times3\), not prime
1
Try \(n=3\): \(2^3+1 = 9\)
2
\(9 = 3 \times 3\), which is not prime
Statement disproved by the counter-example \(n=3\)
Q15 — \(f(x) = (x-2)(2x-1)(x+3)\)
1
\(f(2) = 16+4-26+6 = 0\), confirming \((x-2)\) is a factor
2
Dividing gives \(f(x) = (x-2)(2x^2+5x-3)\)
3
\(2x^2+5x-3 = (2x-1)(x+3)\)
\(f(x) = (x-2)(2x-1)(x+3)\)
Summary
- To simplify an algebraic fraction, factorise the numerator and denominator fully, then cancel any shared factor — never cancel individual terms.
- To add or subtract algebraic fractions, use a common denominator, exactly as with numerical fractions.
- Polynomial division (divide, multiply, subtract, bring down) reduces a cubic to a quotient and remainder — check your answer by expanding back.
- The factor theorem, \(f(a)=0 \iff (x-a)\) is a factor, is the standard route to fully factorising a cubic: find one root by testing values, divide, then factorise the resulting quadratic.
- Choose the right proof method: deduction for a general algebraic argument, exhaustion when only finitely many cases exist, and a single counter-example to disprove a false statement.
This topic builds on GCSE factorising and the early reasoning met in GCSE Sequences & Proof, and the factorising skills from Quadratics. Next in Pure Year 1: Binomial Expansion, which reuses the algebraic rigour built here.
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