Cambridge IGCSE Mathematics 0580: Algebraic fractions and indices II – Study Guide

This guide teaches two subtopics of Topic 2 (Algebra and graphs) in the Cambridge IGCSE Mathematics 0580 syllabus for exams in 2025, 2026 and 2027: 2.4 Indices II and 2.3 Algebraic fractions. Indices II appears in both tiers: Core (C2.4) covers positive, zero and negative indices, and Extended (E2.4) adds fractional indices. Algebraic fractions (E2.3) is Extended only – the syllabus lists no Core content at C2.3. Core candidates sit Papers 1 (non-calculator) and 3 (calculator); Extended candidates sit Papers 2 (non-calculator) and 4 (calculator).

Indices II is the algebra version of Indices I (C1.7/E1.7), which uses numbers such as 7⁻²; for that, see the Number study guide. For the whole of Topic 2 in one place, see the Algebra and graphs study guide and the Algebra and graphs revision notes. Course hub: Cambridge IGCSE Mathematics. Printable topic list: 0580 checklist.

What this unit covers

Syllabus section	What you must be able to do	Tier
C2.4.1 / E2.4.1	Understand and use indices that are positive, zero and negative, including solving equations such as 2ˣ = 32	Core and Extended
E2.4.1	The same with fractional indices, including equations where both sides must be written as powers of one base	Extended only
C2.4.2 / E2.4.2	Understand and use the rules of indices to simplify algebraic expressions	Core and Extended
E2.3.1	Manipulate algebraic fractions: add, subtract, multiply and divide, with numerical or algebraic denominators	Extended only
E2.3.2	Factorise and simplify rational expressions	Extended only

The syllabus notes say knowledge of logarithms is not required for Indices II. Every index equation you meet can be solved by writing both sides as powers of the same number.

Indices II: the rules

An index (power) tells you how many times a base is multiplied by itself: x⁴ = x × x × x × x. These rules hold for any non-zero base.

Rule	In words	Tier
xᵐ × xⁿ = xᵐ⁺ⁿ	Multiply: add the indices	Core and Extended
xᵐ ÷ xⁿ = xᵐ⁻ⁿ	Divide: subtract the indices	Core and Extended
(xᵐ)ⁿ = xᵐⁿ	Power of a power: multiply the indices	Core and Extended
(xy)ⁿ = xⁿyⁿ	A power of a product applies to every factor, numbers included	Core and Extended
x⁰ = 1	Anything (non-zero) to the power 0 is 1	Core and Extended
x⁻ⁿ = 1/xⁿ	A negative index means “one over”	Core and Extended
x^(1/n) = ⁿ√x	A unit fraction index is a root	Extended only
x^(m/n) = (ⁿ√x)ᵐ	Root first, then power	Extended only

Two points cause most trouble:

• The rules only combine powers of the same base. x³ × y² cannot be simplified, and x³ + x² is not x⁵.
• Numbers in front (coefficients) follow ordinary arithmetic. In 4x² × 3x⁵ you multiply 4 × 3 = 12 but add 2 + 5 = 7, giving 12x⁷.

Worked example 1 (Core): simplifying with positive, zero and negative indices

Simplify (a) (2x⁴y)³, (b) 20m⁶n² ÷ 4m⁻³n⁵, (c) 7a⁻²b³ × 3a⁵b⁻⁴.

(a) (2x^4 y)^3 = 2^3 × (x^4)^3 × y^3
               = 8x^12 y^3

(b) 20 ÷ 4 = 5
    m: 6 - (-3) = 9        n: 2 - 5 = -3
    20m^6 n^2 ÷ 4m^-3 n^5 = 5m^9 n^-3 = 5m^9 / n^3

(c) 7 × 3 = 21
    a: -2 + 5 = 3          b: 3 + (-4) = -1
    7a^-2 b^3 × 3a^5 b^-4 = 21a^3 b^-1 = 21a^3 / b

In (a) the 2 is cubed as well – (2x⁴y)³ is not 2x¹²y³. In (b), subtracting a negative index adds it: 6 − (−3) = 9. Both 5m⁹n⁻³ and 5m⁹/n³ are simplified forms.

Also: (5p⁻²)⁰ = 1 (any non-zero p), and (3x⁻²)⁻³ = 3⁻³x⁶ = x⁶/27, because the −3 applies to the 3 too.

Worked example 2 (Core): solving index equations

To solve an equation with the unknown in the index, write both sides as a power of the same base, then set the indices equal.

Solve (a) 3ˣ = 81, (b) 2ˣ = 1/16, (c) 5ˣ⁻¹ = 125.

(a) 81 = 3^4, so 3^x = 3^4  ->  x = 4

(b) 16 = 2^4, so 1/16 = 2^-4
    2^x = 2^-4  ->  x = -4

(c) 125 = 5^3
    5^(x-1) = 5^3  ->  x - 1 = 3  ->  x = 4

In (c), equate the whole index x − 1 with 3, then solve the linear equation. Notice that (b) needs a negative index: a fraction with 1 on top and a power on the bottom is a negative power.

This unit can be examined on the non-calculator papers (1 and 2), so know these by heart: powers of 2 up to 2¹⁰ = 1024, powers of 3 up to 3⁵ = 243, 5³ = 125, 5⁴ = 625, and the squares and cubes in the Number topic.

Fractional indices (Extended only)

A fractional index combines a root and a power. The denominator is the root and the numerator is the power:

x^(1/2) = √x        x^(1/3) = ∛x        x^(m/n) = (ⁿ√x)^m

Take the root first so the numbers stay small: 64^(−2/3) → ∛64 = 4, squared is 16, and the negative index gives 1/16.

Worked example 3 (Extended): simplifying with fractional indices

Simplify (a) (27x⁶)^(2/3), (b) (16x⁸)^(−1/4), (c) (4x²/9)^(−3/2), where x > 0.

(a) 27^(2/3) = (∛27)^2 = 3^2 = 9
    (x^6)^(2/3) = x^(6 × 2/3) = x^4
    (27x^6)^(2/3) = 9x^4

(b) 16^(1/4) = 2, so 16^(-1/4) = 1/2
    (x^8)^(-1/4) = x^-2
    (16x^8)^(-1/4) = (1/2) × x^-2 = 1/(2x^2)

(c) A negative index on a fraction: flip it, then make the index positive
    (4x^2/9)^(-3/2) = (9/(4x^2))^(3/2)
    9^(3/2) = (√9)^3 = 27
    (4x^2)^(3/2) = 4^(3/2) × x^3 = 8x^3
    answer: 27/(8x^3)

The same rules work for mixed fractional indices in products and quotients:

6x^(1/2) × 2x^(-3/2) = 12x^(1/2 - 3/2) = 12x^-1 = 12/x
10x^(3/4) ÷ 5x^(-1/4) = 2x^(3/4 + 1/4) = 2x^1 = 2x

Worked example 4 (Extended): equations needing a common base

When the two sides use different numbers, such as 9 and 27, rewrite both as powers of a smaller base they share (here 3). Then use (aᵐ)ⁿ = aᵐⁿ and equate indices.

Solve (a) 8ˣ = 1/2, (b) 9ˣ⁺² = 27ˣ, (c) (1/25)ˣ = 125.

(a) 8 = 2^3 and 1/2 = 2^-1
    (2^3)^x = 2^-1
    2^(3x) = 2^-1  ->  3x = -1  ->  x = -1/3

(b) 9 = 3^2 and 27 = 3^3
    (3^2)^(x+2) = (3^3)^x
    3^(2x+4) = 3^(3x)
    2x + 4 = 3x  ->  x = 4

(c) 1/25 = 5^-2 and 125 = 5^3
    (5^-2)^x = 5^3
    5^(-2x) = 5^3  ->  -2x = 3  ->  x = -3/2

In (b), (3²)^(x+2) is 3^(2x+4), not 3^(2x+2). Check by substituting: with x = 4, 9⁶ and 27⁴ both equal 531 441.

Algebraic fractions (Extended only)

Algebraic fractions follow exactly the same rules as number fractions. The one new skill is factorising first, using the methods in E2.2: common factors, difference of two squares, and quadratics of the form ax² + bx + c.

Simplifying rational expressions (E2.3.2)

Factorise the numerator and the denominator fully, then cancel common factors – whole brackets or whole terms that multiply everything above and below. You can never cancel a term that is being added or subtracted.

Worked example 5. Simplify (a) (2x² + 5x − 3)/(4x² − 1), (b) (6 − 2x)/(x² − 9), (c) (x³ − 4x)/(x² + 2x).

(a) Numerator:   2x^2 + 5x - 3 = (2x - 1)(x + 3)
    Denominator: 4x^2 - 1 = (2x - 1)(2x + 1)      difference of two squares
    Cancel (2x - 1):   (x + 3)/(2x + 1)

(b) Numerator:   6 - 2x = -2(x - 3)                take out -2 to match the bracket
    Denominator: x^2 - 9 = (x - 3)(x + 3)
    Cancel (x - 3):   -2/(x + 3)

(c) Numerator:   x^3 - 4x = x(x^2 - 4) = x(x - 2)(x + 2)
    Denominator: x^2 + 2x = x(x + 2)
    Cancel x and (x + 2):   x - 2

In (b), spot the reversed bracket: 6 − 2x = −2(x − 3).

To check a simplification, substitute a convenient value. With x = 5 in (a): (50 + 25 − 3)/(100 − 1) = 72/99 = 8/11, and (5 + 3)/(10 + 1) = 8/11. Both agree.

Adding and subtracting (E2.3.1)

Find a common denominator, rewrite each fraction over it, then combine the numerators. Put each numerator in a bracket before you multiply out, so that a minus sign reaches every term.

Worked example 6 (numerical denominators). Write (2x + 1)/4 − (x − 3)/6 as a single fraction.

Lowest common denominator of 4 and 6 is 12.
(2x + 1)/4 = 3(2x + 1)/12        (x - 3)/6 = 2(x - 3)/12

3(2x + 1) - 2(x - 3) = 6x + 3 - 2x + 6 = 4x + 9

Answer: (4x + 9)/12

The −2 multiplies −3 to give +6. Writing −6 here is the most common slip.

Worked example 7 (algebraic denominators). Write as a single fraction (a) x/(x + 4) + 2/(x − 3), (b) 3/(x² − 4) − 1/(x + 2).

(a) Common denominator (x + 4)(x - 3)
    x(x - 3) + 2(x + 4) = x^2 - 3x + 2x + 8 = x^2 - x + 8
    Answer: (x^2 - x + 8)/((x + 4)(x - 3))

(b) x^2 - 4 = (x - 2)(x + 2), which already contains (x + 2).
    Lowest common denominator is (x - 2)(x + 2).
    1/(x + 2) = (x - 2)/((x - 2)(x + 2))
    3 - (x - 2) = 3 - x + 2 = 5 - x
    Answer: (5 - x)/((x - 2)(x + 2))

In (a), x² − x + 8 does not factorise, so nothing cancels and the answer is complete. In (b), factorising first gives the lowest common denominator; multiplying the denominators together gives a cubic and extra cancelling.

Multiplying and dividing (E2.3.1)

To multiply, multiply the numerators and the denominators, cancelling common factors as you go. To divide, turn the second fraction upside down and multiply.

Worked example 8. Simplify (a) (5y/6) × (9/(10y²)), (b) (4x²/15) ÷ (8x/9), (c) ((x² − 4)/(3x)) ÷ ((x + 2)/(6x²)).

(a) (5y × 9)/(6 × 10y^2) = 45y/(60y^2) = 3/(4y)

(b) (4x^2/15) × (9/(8x)) = 36x^2/(120x) = 3x/10

(c) ((x - 2)(x + 2)/(3x)) × (6x^2/(x + 2))
    cancel (x + 2), and 6x^2/(3x) = 2x
    answer: 2x(x - 2)

In (c), factorise x² − 4 before you flip and multiply, so the matching bracket is easy to see.

Common errors

• Multiplying indices when you should add them: x³ × x⁴ = x⁷, not x¹².
• Forgetting to apply a power to the coefficient: (3x²)⁴ = 81x⁸, not 3x⁸.
• Treating a negative index as a negative number: x⁻² = 1/x², not −x².
• Thinking x⁰ = 0 or that 5x⁰ = 1. In fact x⁰ = 1, so 5x⁰ = 5.
• (Extended) Reading x^(2/3) as “square root, cubed”. The denominator is the root: cube root, then square.
• (Extended) Setting indices equal before both sides have the same base, e.g. writing x + 1 = x from 9ˣ⁺¹ = 27ˣ. Rewrite first: 3^(2x+2) = 3^(3x), so x = 2.
• (Extended) Cancelling terms, such as the x² in (x² + 3)/(x² − 1). Only whole factors cancel.
• (Extended) Dropping the bracket around a subtracted numerator, so −(x − 3) becomes −x − 3.
• (Extended) Stopping too soon: after combining fractions, check whether the numerator now shares a factor with the denominator.

Next steps

• Practise with the Algebra and graphs practice questions and the Extended algebra practice, which also covers fractional equations (E2.5).
• Recap the topic with the Algebra and graphs revision notes.
• Find your weak spots with the free 10-minute diagnostic: Core or Extended.
• See other resources for syllabus 0580 on the course hub, and tick off sections on the checklist.

Official syllabus

Cambridge International: Cambridge IGCSE Mathematics 0580 syllabus, for exams in 2025, 2026 and 2027 (Version 3), Topic 2 Algebra and graphs, sections C2.4, E2.3 and E2.4.