AQA GCSE Mathematics 8300: Notation, vocabulary and manipulation – Study Guide

This study guide teaches section 3.2.1 Notation, vocabulary and manipulation (specification references A1 to A7) of the AQA GCSE Mathematics (8300) specification, for teaching from September 2015 with exams from May/June 2017 (version 1.0). The specification splits every reference into three columns: basic foundation content, additional foundation content and Higher content only. Foundation papers can test the first two columns; Higher papers can test all three. Content from the third column is labelled Higher tier only below.

It goes deeper than the topic-level Algebra study guide, which surveys all of Topic 2. Use it with the revision notes for this section and the practice questions for this section. The course hub is AQA GCSE Mathematics, the printable checklist lists every statement, and the free 10-minute diagnostics show where to start.

What this section covers

Ref	What you must be able to do	Tier
A1	Use and interpret algebraic notation: ab, 3y, a², a³, a²b, a/b, fraction coefficients, brackets	Both tiers
A2	Substitute numbers into formulae and expressions, including scientific formulae	Both tiers
A3	Use the words expression, equation, formula, inequality, term, factor	Both tiers
A3	Include identities	Both tiers (additional foundation)
A4	Collect like terms, multiply a single term over a bracket, take out common factors, use the laws of indices	Both tiers
A4	Manipulate expressions with surds, expand two binomials, factorise x² + bx + c including the difference of two squares	Both tiers (additional foundation)
A4	Expand products of two or more binomials, factorise ax² + bx + c, manipulate algebraic fractions	Higher tier only
A5	Use standard formulae and rearrange a formula to change its subject	Both tiers
A6	Know the difference between an equation and an identity; show expressions are equivalent; build arguments with algebra	Both tiers (additional foundation)
A6	Algebraic proof	Higher tier only
A7	Treat simple expressions as functions with inputs and outputs	Both tiers
A7	Inverse and composite functions; f(x), fg(x) and f⁻¹(x) notation	Higher tier only

Any of the three papers can test this section. Paper 1 is non-calculator, so do every manipulation by hand; on Papers 2 and 3 a calculator helps only with arithmetic after substituting.

A1: Algebraic notation

• ab means a × b. Write the number first: 7b, not b7.
• 3y means y + y + y, and also 3 × y.
• a² means a × a, a³ means a × a × a, and a²b means a × a × b.
• a/b means a ÷ b.
• Coefficients are written as fractions, not decimals: write ½x or x/2, not 0.5x.
• Brackets group terms: 2(a + b) means 2 × (a + b).

The specification adds that answers are expected in simplest form even when the question does not say so. So 2y + 3y left unsimplified can cost the final mark.

Worked example. Write each in simplest algebraic notation: (a) m × m × m × n, (b) p ÷ 4, (c) 0.5 × x, (d) y + y + y + y.

(a) m × m × m × n = m³n
(b) p ÷ 4         = p/4
(c) 0.5 × x       = ½x   (or x/2)
(d) y + y + y + y = 4y

A2: Substitution

To substitute, replace each letter with its value in brackets, then follow the order of operations. Brackets matter most with negative numbers and powers.

The specification includes scientific formulae. If a formula is unfamiliar, the question will give it to you. The kinematics formulae (v = u + at, s = ut + ½at², v² = u² + 2as) are on the specification’s list of formulae that are given in the exam.

Worked example 1. Use s = ut + ½at² to find s when u = 3, t = 4 and a = −2.

s = (3)(4) + ½(−2)(4)²
  = 12 + ½(−2)(16)
  = 12 − 16
  = −4

Worked example 2. Find the value of 2x² − 5x when x = −3.

2(−3)² − 5(−3) = 2(9) + 15 = 18 + 15 = 33

The common slip is working out −3² as −9. In brackets, (−3)² = 9.

A3: Vocabulary

Word	Meaning	Example
Term	One part of an expression, separated by + or −	4p and −7 in 4p − 7
Expression	Terms combined, with no equals sign	4p − 7
Equation	Two sides equal for particular values only	4p − 7 = 9 (true only for p = 4)
Formula	A rule linking two or more variables	P = 2l + 2w
Inequality	Two sides compared with <, >, ≤ or ≥	4p − 7 > 9
Identity	Two sides equal for every value; written with ≡	4(p − 2) ≡ 4p − 8
Factor	Something that divides exactly into an expression	3 and (x + 2) are factors of 3(x + 2)

A4: Simplifying and manipulating expressions

Collecting like terms

Like terms have exactly the same letters and powers. mn and m are not like terms.

7m − 2n + 3mn − 4m + 5n − mn
= (7m − 4m) + (−2n + 5n) + (3mn − mn)
= 3m + 3n + 2mn

Multiplying a single term over a bracket

Multiply every term inside by the term outside, keeping signs.

3x(2x − 5) = 6x² − 15x
−4(y − 3)  = −4y + 12

Taking out common factors

“Factorise fully” means take out the highest common factor: the largest number and every letter that divides all terms.

12a²b + 18ab²
HCF of 12 and 18 = 6; both terms contain a and b
= 6ab(2a + 3b)

Check by expanding back.

The laws of indices

aᵐ × aⁿ = aᵐ⁺ⁿ        aᵐ ÷ aⁿ = aᵐ⁻ⁿ        (aᵐ)ⁿ = aᵐⁿ

Work out the numbers and the powers separately.

(2p³)⁴ = 2⁴ × p¹² = 16p¹²
15x⁷y² ÷ 5x³y = 3x⁴y

Zero and negative integer indices (a⁰ = 1, a⁻ⁿ = 1/aⁿ) come from N7 in the Number section. Fractional indices are Higher tier only (N7).

Expanding two binomials

Multiply each term in the first bracket by each term in the second, then collect like terms.

(x + 7)(x − 4) = x² − 4x + 7x − 28 = x² + 3x − 28
(2x − 3)²      = (2x − 3)(2x − 3) = 4x² − 6x − 6x + 9 = 4x² − 12x + 9

(2x − 3)² is not 4x² + 9. Always write a squared bracket out twice.

Expressions involving surds

The same rules work with surds. The fact you need is √5 × √5 = 5.

(3 + √5)(2 − √5) = 6 − 3√5 + 2√5 − 5 = 1 − √5

Factorising x² + bx + c

Find two numbers that multiply to c and add to b.

x² − 2x − 24: multiply to −24, add to −2 → −6 and +4
= (x − 6)(x + 4)

The difference of two squares is a special case with no x term:

a² − b² = (a + b)(a − b)
y² − 49 = (y + 7)(y − 7)

Higher tier only: products of more than two binomials

Expand two brackets first, then multiply the result by the third.

(x + 2)(x − 1)(x + 3)
(x + 2)(x − 1) = x² + x − 2
(x² + x − 2)(x + 3) = x³ + 3x² + x² + 3x − 2x − 6
                    = x³ + 4x² + x − 6

Higher tier only: factorising ax² + bx + c

Multiply a × c, find two numbers that multiply to ac and add to b, split the middle term, then factorise in pairs.

3x² + 10x − 8:  ac = 3 × (−8) = −24
Numbers: 12 and −2 (product −24, sum 10)
3x² + 12x − 2x − 8 = 3x(x + 4) − 2(x + 4) = (3x − 2)(x + 4)

Higher tier only: algebraic fractions

To simplify, factorise top and bottom fully, then cancel whole brackets. Never cancel single terms.

(x² − 9)/(x² + x − 6) = (x + 3)(x − 3) / ((x + 3)(x − 2)) = (x − 3)/(x − 2)

To add or subtract, use a common denominator, exactly as with number fractions.

2/(x + 1) + 3/(x − 2)
= [2(x − 2) + 3(x + 1)] / ((x + 1)(x − 2))
= (2x − 4 + 3x + 3) / ((x + 1)(x − 2))
= (5x − 1) / ((x + 1)(x − 2))

A5: Using and rearranging formulae

The specification includes formulae from other subjects, given in words or in symbols. A formula in words can be turned into symbols first: “total cost in pounds = 15 + 3 × number of hours” becomes C = 15 + 3h.

To change the subject, undo the operations on the new subject in reverse order, doing the same to both sides.

Worked example 1. Make h the subject of C = 15 + 3h.

C − 15 = 3h
h = (C − 15)/3

Worked example 2. Make x the subject of y = (3x − 2)/5.

5y = 3x − 2
5y + 2 = 3x
x = (5y + 2)/3

Worked example 3. Make r the subject of V = πr²h.

V/(πh) = r²
r = √(V/(πh))

Some questions put the new subject in two terms. The method is to expand, gather those terms on one side, then factorise. (The specification does not split A5 by tier.)

Make t the subject of a(t + 2) = bt − 1
at + 2a = bt − 1
2a + 1 = bt − at
2a + 1 = t(b − a)
t = (2a + 1)/(b − a)

A6: Equations, identities and algebraic argument

An equation is true for particular values. 3(x + 4) = 2x + 1 is true only when x = −11. An identity is true for every value: 3(x + 4) − x ≡ 2x + 12, because the left side simplifies to exactly the right side.

To show that two expressions are equivalent, start from one side and simplify until it matches the other. Show every line.

Worked example. Show that (n + 3)² − (n − 3)² ≡ 12n.

(n + 3)² = n² + 6n + 9
(n − 3)² = n² − 6n + 9
(n² + 6n + 9) − (n² − 6n + 9) = 12n  ✓

Using algebra to support an argument. Show that the sum of any three consecutive whole numbers is a multiple of 3.

Let the numbers be n, n + 1, n + 2
Sum = 3n + 3 = 3(n + 1)
3(n + 1) has a factor of 3, so the sum is a multiple of 3.

Higher tier only: proof

A proof must work for every case, so testing examples never counts. Useful forms: an even number is 2n, an odd number is 2n + 1, and consecutive odd numbers are 2n − 1 and 2n + 1.

Worked example. Prove that the difference between the squares of two odd numbers that differ by 4 is a multiple of 8.

Let the odd numbers be 2n − 1 and 2n + 3
(2n + 3)² − (2n − 1)²
= (4n² + 12n + 9) − (4n² − 4n + 1)
= 16n + 8 = 8(2n + 1)
2n + 1 is an integer, so the difference is a multiple of 8.

The final sentence, linking the algebra to the claim, is part of the proof.

A7: Functions

At both tiers, a simple expression can be read as a function: an input goes in, a rule is applied, an output comes out. A function machine “× 3, then − 5” turns input x into output 3x − 5, so input 6 gives output 13.

Higher tier only: f(x), f⁻¹(x) and fg(x)

Let f(x) = 3x − 5 and g(x) = x² + 1.

• f(4) means substitute 4: f(4) = 12 − 5 = 7.
• f⁻¹(x) is the inverse function, the reverse process. Write y = 3x − 5 and make x the subject: x = (y + 5)/3, so f⁻¹(x) = (x + 5)/3. Check: f⁻¹(7) = 4.
• fg(x) is a composite function: do g first, then f. fg(x) = f(x² + 1) = 3(x² + 1) − 5 = 3x² − 2.
• gf(x) is different: gf(x) = g(3x − 5) = (3x − 5)² + 1 = 9x² − 30x + 26.

For a number, work from the inside out: fg(2) = f(5) = 10, but gf(2) = g(1) = 2.

Common errors

• Writing 0.5x instead of ½x, or leaving 2 × a × b instead of 2ab.
• Substituting a negative without brackets, so −3² becomes −9.
• Adding powers when raising to a power: (p³)⁴ is p¹², not p⁷.
• Writing (2x − 3)² as 4x² + 9 and losing the middle term.
• Taking out a common factor that is not the highest, so “factorise fully” loses a mark.
• Cancelling single terms in an algebraic fraction, such as crossing out x² from top and bottom.
• Treating fg(x) as “f first”. The function nearest the x acts first.
• Using numerical examples as a “proof”.

Next steps

Condense this into the revision notes, then test yourself with the practice questions. For the rest of Topic 2, see the Algebra revision notes and Algebra practice questions.

Official syllabus

AQA GCSE Mathematics (8300) specification, for teaching from September 2015, exams from May/June 2017, version 1.0, published by AQA – section 3.2.1 Notation, vocabulary and manipulation (A1 to A7) and Appendix: mathematical formulae.