Edexcel International GCSE Mathematics A 4MA1: Use of symbols and algebraic manipulation – Study Guide

This study guide teaches sections 2.1 Use of symbols and 2.2 Algebraic manipulation from Topic 2, Equations, formulae and identities, of the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for the January and June series examined on that specification. 4MA1 is tiered: Foundation papers 1F and 2F, Higher papers 1H and 2H, each 2 hours and 100 marks, and a calculator may be used on all of them. Both tiers have content under 2.1 and 2.2; the extra Higher Tier statements are labelled Higher tier only below, and Higher questions assume all the Foundation content.

When you have worked through this guide, test your recall with the revision notes and then try the practice questions. The Edexcel IGCSE Mathematics course hub links the rest of the course, and the printable checklist lists every specification statement.

There is also an earlier topic-level guide to use of symbols and algebraic manipulation, with its own revision notes and practice set. The examples here are different from theirs.

What this unit covers

Statement	What you must be able to do	Tier
2.1A	Understand that symbols can stand for numbers in equations, or variables in expressions and formulae	Both
2.1B	Understand that algebraic expressions follow the generalised rules of arithmetic	Both
2.1C	Use index notation for positive and negative integer powers, including zero	Both
2.1D	Use the index laws in simple cases	Both
2.1A (Higher)	Use index notation with fractional, negative and zero powers	Higher tier only
2.2A	Evaluate expressions by substituting numbers for letters	Both
2.2B	Collect like terms	Both
2.2C	Multiply a single term over a bracket	Both
2.2D	Take out common factors (factorise fully)	Both
2.2E	Expand the product of two simple linear expressions	Both
2.2F	Factorise quadratic expressions of the form x² + bx + c	Both
2.2A (Higher)	Expand the product of two or more linear expressions	Higher tier only
2.2B (Higher)	Factorise quadratic expressions in general	Higher tier only
2.2C (Higher)	Manipulate algebraic fractions with numeric, linear or quadratic numerators and denominators	Higher tier only
2.2D (Higher)	Complete the square for a given quadratic	Higher tier only
2.2E (Higher)	Use algebra to support and construct proofs	Higher tier only

2.1 Use of symbols

Letters, expressions and formulae

A letter can stand for a particular unknown number, as in the equation 3x + 1 = 13, where x can only be 4. A letter can also stand for a variable, which takes many values, as in the expression 3x + 1 or the formula A = πr².

• An expression has no equals sign: 4a − 7b.
• An equation is true only for particular values: 4a − 7 = 21.
• A formula links variables: v = u + at.

The rules of arithmetic still apply

Algebra follows the same rules as arithmetic, written with letters:

• a + b = b + a and ab = ba (order does not matter for adding or multiplying)
• (a + b) + c = a + (b + c) and (ab)c = a(bc)
• a(b + c) = ab + ac (multiplying over a bracket)
• a − b is not b − a, and a ÷ b is not b ÷ a

So 3 × y × 2 × y can be reordered to 3 × 2 × y × y = 6y².

Index notation and the index laws

aⁿ means n copies of a multiplied together: a × a × a = a³. The laws, for the same base a:

Law	Rule	Example
Multiply	aᵐ × aⁿ = aᵐ⁺ⁿ	x² × x⁵ = x⁷
Divide	aᵐ ÷ aⁿ = aᵐ⁻ⁿ	x⁹ ÷ x⁴ = x⁵
Power of a power	(aᵐ)ⁿ = aᵐⁿ	(x³)⁴ = x¹²
Zero power	a⁰ = 1	7⁰ = 1
Negative power	a⁻ⁿ = 1/aⁿ	a⁻⁵ = 1/a⁵

The zero rule follows from the divide law: a³ ÷ a³ = a⁰, and anything divided by itself is 1.

Worked example 1 (both tiers). Simplify (a) y⁷ ÷ y⁻² (b) (p⁻³)² (c) 3x⁴ × 5x⁻⁴ (d) (2a³)² × a⁻⁴

(a) y^7 ÷ y^(-2) = y^(7 - (-2)) = y^9
(b) (p^(-3))^2  = p^(-6) = 1/p^6
(c) 3x^4 × 5x^(-4) = (3 × 5) × x^(4 + (-4)) = 15x^0 = 15
(d) (2a^3)^2 = 2^2 × a^6 = 4a^6
    4a^6 × a^(-4) = 4a^2

In (d) the 2 inside the bracket is squared too. Writing 2a⁶ is a common slip.

Fractional powers – Higher tier only

A fractional power means a root:

• a^(1/n) is the nth root of a, so 9^(1/2) = √9 = 3 and 64^(1/3) = ∛64 = 4
• a^(m/n) = (ⁿ√a)ᵐ: take the root first, then the power, so the numbers stay small
• a negative fractional power is the reciprocal: a^(−m/n) = 1/a^(m/n)

Worked example 2 (Higher). Work out (a) 32^(3/5) (b) (8/27)^(−2/3). (c) Simplify (16x⁸)^(3/4).

(a) fifth root of 32 = 2,  then 2^3 = 8
(b) negative power: flip the fraction  (27/8)^(2/3)
    cube root of 27/8 = 3/2,  then (3/2)^2 = 9/4
(c) 16^(3/4): fourth root of 16 = 2, 2^3 = 8
    (x^8)^(3/4) = x^(8 × 3/4) = x^6
    answer 8x^6

2.2 Algebraic manipulation (both tiers)

Substitution

Replace each letter with its value in brackets, then follow the order of operations (powers before multiplying).

Worked example 3. Find the value of 3a² − 2ab when (i) a = −4, b = 2.5 (ii) a = 1/2, b = −3.

(i)  3(-4)^2 - 2(-4)(2.5) = 3(16) - (-20) = 48 + 20 = 68
(ii) 3(1/2)^2 - 2(1/2)(-3) = 3/4 - (-3) = 3/4 + 3 = 15/4

Without brackets, a calculator reads −4² as −16.

Collecting like terms

Like terms have exactly the same letters with the same powers. x²y and xy² are not like terms.

5x²y − 3xy² + 2x²y + xy² = 7x²y − 2xy²

One bracket, and taking out common factors

Expanding and factorising are opposite processes.

• Expand: 4p(3p − 2q) = 12p² − 8pq
• Factorise fully: 18a²b − 12ab² = 6ab(3a − 2b)

“Fully” means the highest common factor of the numbers and the letters. 6a(3ab − 2b²) or 3ab(6a − 4b) are only partly factorised and lose the final mark.

Expanding two linear brackets

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

(x − 7)(x + 3) = x² + 3x − 7x − 21 = x² − 4x − 21

Factorising x² + bx + c

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

Worked example 4. Factorise (a) x² − 11x + 24 (b) x² + 2x − 35.

(a) product +24, sum -11: both negative -> -3 and -8
    x^2 - 11x + 24 = (x - 3)(x - 8)
(b) product -35, sum +2: signs differ, bigger one positive -> +7 and -5
    x^2 + 2x - 35 = (x + 7)(x - 5)

Check by expanding: (x + 7)(x − 5) = x² − 5x + 7x − 35 = x² + 2x − 35.

2.2 Algebraic manipulation – Higher tier only

Expanding three or more linear brackets

Multiply two brackets, simplify, then multiply by the next one.

Worked example 5. Expand and simplify (2x − 1)(x + 3)(x − 4).

(2x - 1)(x + 3) = 2x^2 + 6x - x - 3 = 2x^2 + 5x - 3
(2x^2 + 5x - 3)(x - 4)
  = 2x^3 - 8x^2 + 5x^2 - 20x - 3x + 12
  = 2x^3 - 3x^2 - 23x + 12

Quick check: put x = 2. The original gives (3)(5)(−2) = −30. The answer gives 16 − 12 − 46 + 12 = −30.

Factorising any quadratic

For ax² + bx + c with a ≠ 1, find two numbers that multiply to ac and add to b, split the middle term, and factorise in pairs.

Worked example 6. Factorise 10x² − 11x − 6.

ac = 10 × (-6) = -60; need sum -11  ->  -15 and +4
10x^2 - 15x + 4x - 6
= 5x(2x - 3) + 2(2x - 3)
= (2x - 3)(5x + 2)

Always look for a common factor first, then for the difference of two squares, p² − q² = (p − q)(p + q):

• 9x² − 49y² = (3x − 7y)(3x + 7y)
• 2x² − 50 = 2(x² − 25) = 2(x − 5)(x + 5)

Algebraic fractions

The rules are the rules of ordinary fractions. To simplify, factorise top and bottom fully, then cancel whole common factors (brackets), never single terms.

Worked example 7. Simplify (x² − 9)/(x² + x − 6).

x^2 - 9     = (x - 3)(x + 3)
x^2 + x - 6 = (x + 3)(x - 2)
cancel (x + 3):  (x - 3)/(x - 2)

To add or subtract, use a common denominator. Denominators can be numbers or brackets.

Worked example 8. Write as single fractions (a) (x + 2)/3 − (x − 1)/4 (b) 3/(x + 4) + 2/(x − 1).

(a) common denominator 12
    [4(x + 2) - 3(x - 1)] / 12 = (4x + 8 - 3x + 3)/12 = (x + 11)/12

(b) common denominator (x + 4)(x - 1)
    [3(x - 1) + 2(x + 4)] / [(x + 4)(x - 1)]
    = (3x - 3 + 2x + 8) / [(x + 4)(x - 1)]
    = (5x + 5) / [(x + 4)(x - 1)]
    = 5(x + 1) / [(x + 4)(x - 1)]

In (a) the bracket around the second numerator matters: −3(x − 1) gives +3, not −3. To divide by a fraction, multiply by its reciprocal, then factorise and cancel as before.

Completing the square

For x² + bx + c, halve b: x² + bx + c = (x + b/2)² − (b/2)² + c. If the x² coefficient is not 1, take it out of the x terms first.

Worked example 9. Write (a) x² − 8x + 3 in the form (x + p)² + q (b) 5x² + 20x + 7 in the form a(x + b)² + c.

(a) half of -8 is -4
    (x - 4)^2 = x^2 - 8x + 16, so subtract 16
    x^2 - 8x + 3 = (x - 4)^2 - 16 + 3 = (x - 4)^2 - 13

(b) 5x^2 + 20x + 7 = 5(x^2 + 4x) + 7
                   = 5[(x + 2)^2 - 4] + 7
                   = 5(x + 2)^2 - 20 + 7
                   = 5(x + 2)^2 - 13

So a = 5, b = 2, c = −13. Since (x + 2)² is never negative, the least value of 5x² + 20x + 7 is −13, when x = −2.

Algebraic proof

A proof must work for every value, so use letters, not examples. Useful forms, for any integer n: even number 2n; odd number 2n + 1; consecutive integers n, n + 1; consecutive odd numbers 2n − 1, 2n + 1. To show something is a multiple of k, get it into the form k × (an integer).

Worked example 10. Prove that the difference between the squares of two consecutive odd numbers is always a multiple of 8.

Let the odd numbers be 2n - 1 and 2n + 1, where n is an integer.
(2n + 1)^2 - (2n - 1)^2
= (4n^2 + 4n + 1) - (4n^2 - 4n + 1)
= 8n
8n = 8 × n, and n is an integer, so the difference is a multiple of 8.

Testing 3 and 5, then 7 and 9, is not a proof. Finish with a sentence that links your algebra to the claim.

Calculators and checking

A calculator is allowed on every 4MA1 paper, but calculators with built-in symbolic algebra are prohibited, so all manipulation is done by hand. Use your calculator to check: substitute a simple value such as x = 2 into the question and into your answer. If the two numbers differ, there is an error.

Common errors

• Adding powers of different bases: x² × y³ stays x²y³.
• Treating a⁰ as 0, or a⁻² as −a².
• Writing (x + 4)² as x² + 16. The middle term 8x is missing.
• Subtracting only the first term of a second bracket: −(x − 6) is −x + 6.
• Cancelling x in (x + 3)/(x + 5). Only whole factors cancel.
• Completing the square on 3x² + … without taking 3 out first, or forgetting to multiply the “subtract” term by 3.
• In a proof, using n and n + 2 for consecutive odd numbers without stating that n is odd.

Next steps

Go to the revision notes and practice questions linked at the top. To find gaps elsewhere in the course, use our free diagnostics.

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification, Issue 2, November 2017, Pearson Education Limited (first assessment June 2018). Topic 2, Equations, formulae and identities: sections 2.1 Use of symbols and 2.2 Algebraic manipulation (Foundation and Higher Tier content).