Pearson Edexcel IGCSE Mathematics: Use of Symbols and Algebraic Manipulation — Revision Notes

Condensed for the final weeks. For the full explanation, use the Use of Symbols and Algebraic Manipulation study guide.

Index notation (2.1)

x^(-n) = 1 / x^n           (negative power = reciprocal)
x^0 = 1                    (for any non-zero base)
x^(1/n) = nth root of x    (fractional power = a root)

Treat these as three separate rules, not one blurred idea of “powers can look different.” Index laws for combining powers only apply to the same base.

Expanding three linear expressions (2.2A)

Systematic method: multiply the first two brackets fully, simplify, then multiply by the third bracket — don’t attempt all three at once, where arithmetic slips creep in.

Factorising quadratics (2.2B)

Factorise when the quadratic has clean integer roots. Example: 6x² − 5x − 6.

Completing the square (2.2D)

Use when a quadratic does not factorise cleanly, or when finding a turning point.

Worked example: write 2x² + 6x − 1 in the form a(x + b)² + c.

2x^2 + 6x - 1
= 2(x^2 + 3x) - 1              [factor out coefficient of x^2]
= 2[(x + 1.5)^2 - 2.25] - 1    [complete the square inside]
= 2(x + 1.5)^2 - 4.5 - 1       [multiply -2.25 by 2 !]
= 2(x + 1.5)^2 - 5.5

a = 2, b = 1.5, c = -5.5

The single most common error: forgetting to multiply the adjustment term (−2.25) by the factored-out coefficient (2) before combining with the constant already outside the bracket.

Algebraic fractions (2.2C)

Combines several skills at once: factorise → cancel common factors → find a common denominator. Revise this only after 2.2B is genuinely fluent — a factorising weakness will surface here.

Worked example: express (3x+1)/(x+2) − (x−2)/(x−1) as a single fraction.

Common denominator: (x + 2)(x - 1)

(3x + 1)(x - 1) - (x - 2)(x + 2)
---------------------------------
        (x + 2)(x - 1)

Numerator: (3x^2 - 2x - 1) - (x^2 - 4) = 2x^2 - 2x + 3

Result: (2x^2 - 2x + 3) / [(x + 2)(x - 1)]

Two commonly rushed steps: expanding each bracket fully first, and subtracting the entire second numerator, not just its first term.

Proof using algebra (2.2E)

Algebra is used to construct proofs — express an unknown generally (e.g. “let the number be n” or “2n” for any even number), manipulate, and show the required result holds for all valid values, not just checked examples.

Worked example: factorising vs. completing the square

Decide the more efficient method for x^2 + 6x + 5, then for x^2 + 4x + 1.

x^2 + 6x + 5:
  Look for two numbers multiplying to 5, adding to 6: 1 and 5.
  Factorises cleanly: (x + 1)(x + 5).
  -> FACTORISING is the faster method here.

x^2 + 4x + 1:
  No integer pair multiplies to 1 and adds to 4.
  Does not factorise with integers.
  -> COMPLETING THE SQUARE is required:
     x^2 + 4x + 1 = (x + 2)^2 - 4 + 1 = (x + 2)^2 - 3

Practising this decision – try factorising first, and switch to completing the square only when it doesn’t yield clean integers – rather than defaulting to one method regardless of the expression, saves time in an exam and avoids forcing a factorisation that doesn’t actually exist.

Where this content leads

Sub-topics 2.1 and 2.2 are the toolkit every later sub-topic in Topic 2 (Equations, Formulae and Identities) depends on: solving quadratic equations by factorisation, the quadratic formula and completing the square (2.7) uses factorising and completing the square directly; functions and graphs, in Topic 3 (Sequences, Functions and Graphs), require confident algebraic manipulation to find intercepts, turning points and asymptotes. Treat fluency here as a prerequisite to secure before moving on, not content to revise once in isolation.

Key terms

Index notation — writing repeated multiplication as a power (base and exponent). Factorising — writing an expression as a product of factors. Completing the square — rewriting a quadratic as a(x+b)²+c. Algebraic fraction — a fraction with an algebraic expression in the numerator and/or denominator.

A note on proof by algebra (2.2E)

A typical exam task: “Prove that the sum of any two consecutive odd numbers is always even.” Represent the first odd number generally as 2n + 1 (for any integer n), so the next consecutive odd number is 2n + 3. Their sum is (2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1), which is a multiple of 4 and therefore always even, for every integer value of n – not just the specific examples you might check by hand. Representing the unknown generally, rather than proving the claim only for a few chosen numbers, is what separates a genuine algebraic proof from an unconvincing pattern-spotting answer.

Common mistakes

• Applying an index law to terms with different bases.
• Sign errors when expanding three linear expressions, especially with a negative term in more than one bracket.
• Cancelling additive terms in an algebraic fraction as though they were multiplicative factors — not valid.
• Forgetting to multiply the adjustment term by the factored-out coefficient when completing the square.

Quick self-test

1. Simplify x⁻³, x⁰, and x^(1/2) in words.
2. Expand (x + 1)(x + 2)(x − 3).
3. Write 3x² − 12x + 5 in completed-square form.
4. Simplify (2x² + 3x)/(4x² − 9).
5. Express (x+1)/(x−1) + (x−2)/(x+1) as a single fraction.

Answers:

1. x⁻³ is the reciprocal of x cubed, 1/x³; x⁰ is 1 for any non-zero x; x^(1/2) is the square root of x, √x.
2. (x + 1)(x + 2)(x − 3) = (x² + 3x + 2)(x − 3) = x³ − 7x − 6.
3. 3x² − 12x + 5 = 3(x² − 4x) + 5 = 3[(x − 2)² − 4] + 5 = 3(x − 2)² − 7.
4. (2x² + 3x)/(4x² − 9) = x(2x + 3) / [(2x − 3)(2x + 3)] = x/(2x − 3).
5. (x+1)/(x−1) + (x−2)/(x+1) = [(x+1)² + (x−2)(x−1)] / [(x−1)(x+1)] = (x² + 2x + 1 + x² − 3x + 2) / (x² − 1) = (2x² − x + 3)/(x² − 1).

Higher-tier candidates should expect 2.2A–2.2E assessed both in isolation and combined within a single longer question — practising them together is genuinely representative of exam demand.

Official syllabus

Pearson Edexcel International GCSE Mathematics (Specification A) (4MA1) specification, Issue 2, November 2017 — qualifications.pearson.com.