Sec 3 E-Math: Algebraic Expressions & Formulae

Overview

What you’ll learn

Simplify linear expressions, removing brackets and collecting like terms.
Expand products of algebraic expressions, including the identities (a + b)², (a − b)² and a² − b².
Factorise linear expressions by grouping and quadratic expressions of the form ax² + bx + c.
Multiply and divide simple algebraic fractions, and add or subtract fractions with linear or quadratic denominators.
Find the nth term of a number pattern and use it to predict later terms.
Change the subject of a formula and evaluate formulae, including with squares and roots.

In 30 seconds

1Collect like terms only — 3x and 5x combine to 8x, but 3x and 5x² never merge.
2A minus in front of a bracket flips every sign inside: −2(3x − 5) = −6x + 10.
3The three special identities: a² − b² = (a + b)(a − b), (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b².
4Factorise quadratics by finding two numbers with the right product and sum; when the x² coefficient is not 1, split the middle term or use factor pairs of ac.
5Algebraic fractions follow number-fraction rules: factorise every denominator first, then cancel whole factors only.
6An nth-term rule turns a pattern into a formula; check it against two figures before trusting it.
7Changing the subject undoes operations in reverse order — roots come out last.
8Substitute after simplifying where possible, and keep negatives in brackets when squaring.

Notes · 1 of 5

Simplifying expressions

Expand brackets, collect like terms, and respect the sign in front of every bracket.

An expression is a recipe of terms. Like terms share exactly the same letter part: 3x and −5x are alike, 3x and 3x² are not. Simplifying means expanding where needed and collecting like terms — the answer is still an expression, not an equation, so there is nothing to solve.

1Expand every bracket
3(2x − 1) − 2(x − 4) = 6x − 3 − 2x + 8. The −2 flips both signs inside the second bracket.
2Collect like terms
6x − 2x = 4x and −3 + 8 = 5.
3Write the result
The expression simplifies to 4x + 5.

Notes · 2 of 5

Expansion and the special identities

Every product expands term by term; three patterns are worth memorising outright.

The three special identities (printed in the formula list)
Name	Identity	Example
Square of a sum	(a + b)² = a² + 2ab + b²	(x + 5)² = x² + 10x + 25
Square of a difference	(a − b)² = a² − 2ab + b²	(2x − 3)² = 4x² − 12x + 9
Difference of two squares	a² − b² = (a + b)(a − b)	x² − 9 = (x + 3)(x − 3)

The middle term is where marks go missing: (a + b)² is a² + 2ab + b², not a² + b². A quick numerical test settles any doubt — (3 + 4)² = 49 while 3² + 4² = 25, so the cross term is real. The difference of two squares runs both ways: expanding (x + 3)(x − 3) gives x² − 9, and factorising x² − 9 reverses it.

Notes · 3 of 5

Factorisation: common factors, grouping, quadratics

Factorising is expansion reversed: hunt for shared factors, then factor pairs.

1Take out the highest common factor
6x²y − 9xy² = 3xy(2x − 3y). Every term must shrink by the factor you pull out.
2Group when four terms appear
3x + 6y + ax + 2ay = 3(x + 2y) + a(x + 2y) = (x + 2y)(a + 3) .
3Spot the identities
x² − 8x + 16 = (x − 4)² and 4x² − 25 = (2x + 5)(2x − 5) — always check for a² ± 2ab + b² and a² − b² before anything longer.

For x² + bx + c the product–sum search uses c directly: 3x² − 5x − 2 needs product −6 and sum −5, namely −6 and +1, giving (3x + 1)(x − 2). Sign rules: when c is negative the two numbers have opposite signs; when c is positive and b is negative, both are negative.

Notes · 4 of 5

Algebraic fractions

Factorise every denominator, cancel whole factors, and use a common denominator to add or subtract.

Multiplication multiplies numerators and denominators (factorised form makes cancelling visible); division multiplies by the reciprocal. Addition and subtraction need a common denominator — which is easiest to see once every denominator is factorised.

1Multiply
(3a/4b) × (5ab/3) = 15a²b/12b = 5a²/4 after cancelling b and 3.
2Divide
(3x/4) ÷ (9x²/10) = (3x/4) × (10/9x²) = 30x/36x² = 5/(6x) .
3Add
1/(x² − 9) + 2/(x − 3): factorise x² − 9 = (x − 3)(x + 3), so the LCM is (x − 3)(x + 3).

Notes · 5 of 5

Patterns, nth term, and changing the subject

Turn a pattern into an nth-term rule, then bend formulae around to make any letter the subject.

A linear pattern grows by the same step: 5, 8, 11, … rises by 3, so the rule is 3n + 2 (the constant tops the first term up from 3 to 5). Figure-type patterns work the same way: a row of squares made of matchsticks uses 4, 7, 10, … sticks — each new square adds 3, so the nth figure needs 3n + 1 sticks. Always check the rule against a second term before using it.

1Linear subjects
v = u + at gives t = (v − u)/a: subtract first, then divide.
2Bracketed subjects
P = 2(l + b) gives b = P/2 − l: divide by 2 first (or expand, then isolate).
3Squared subjects
Take the square root only after everything else is clear of the letter.

Revise

Key terms

The words examiners expect you to use precisely.

Like terms: Terms with exactly the same letter part, such as 4x and −7x, which can be collected.
Coefficient: The number multiplying a letter part; in −5x² the coefficient is −5.
Expand: Multiply out brackets so no bracketed group remains.
Factorise: Write an expression as a product of factors — the reverse of expanding.
Difference of two squares: The identity a² − b² = (a + b)(a − b), recognisable by two perfect squares separated by a minus.
Perfect square: An expression matching (a ± b)² = a² ± 2ab + b².

Show 4 more key terms

Algebraic fraction: A fraction whose numerator, denominator, or both contain letters.
Lowest common denominator: The LCM of fraction denominators, found after factorising each one.
nth term: A rule in n giving every term of a sequence; the 100th term needs no list.
Subject of a formula: The single letter isolated on one side of a formula; rearranging gives a new subject.

Revise

Common mistakes

Where students most often lose marks in this chapter — and what to do instead.

Mistake: Collecting unlike terms, e.g. 3x + 5x² = 8x³.
The letter parts differ; powers do not add across different terms.
Instead: 3x + 5x² stays as it is — only like terms combine.
Mistake: Not flipping every sign when a minus precedes a bracket: 5 − (2x − 3) = 5 − 2x − 3.
The minus multiplies the whole bracket by −1.
Instead: 5 − (2x − 3) = 5 − 2x + 3 = 8 − x.

Show 10 more mistakes

Mistake: Writing (a + b)² = a² + b².
The middle term 2ab is missing: (3 + 4)² = 49, not 25.
Instead: Square the bracket in full or use the identity with the cross term.
Mistake: Taking √(x² + 9) as x + 3.
A root does not split over addition — test x = 4: √25 = 5, not 7.
Instead: Only products split under a root; x² + 9 does not factorise.
Mistake: Cancelling a single term across a fraction: (x² + 9)/(x + 3) → x + 3?
Only whole factors common to numerator and denominator cancel.
Instead: Factorise first: (x² − 9)/(x + 3) = x − 3, but (x² + 9)/(x + 3) cannot be simplified.
Mistake: Forgetting to flip the second fraction when dividing: (3x/4) ÷ (9x²/10) treated as (3x/4) × (9x²/10).
Division by a fraction multiplies by its reciprocal.
Instead: (3x/4) × (10/9x²) = 5/(6x).
Mistake: Adding fractions without a common denominator: 1/(x − 2) + 1/(x − 3) written as 2/(2x − 5).
Numerators and denominators do not add separately.
Instead: LCM (x − 2)(x − 3): the sum is (2x − 5)/((x − 2)(x − 3)).
Mistake: Sign errors in factorising with a negative c, e.g. 3x² − 5x − 2 = (3x − 1)(x + 2).
That expansion gives 3x² + 5x − 2 — the middle term has the wrong sign.
Instead: Product −6 and sum −5 need −6 and +1: (3x + 1)(x − 2).
Mistake: Partially factorising, e.g. stopping at 3(x + 2y) + a(x + 2y).
The common bracket (x + 2y) must still be pulled out.
Instead: Grouping finishes as (x + 2y)(3 + a).
Mistake: Changing the subject by moving terms across without operating on both sides, e.g. v = u + at giving at = v (dropping u).
Each rearrangement step applies to the whole equation.
Instead: Subtract u from both sides first: at = v − u, then t = (v − u)/a.
Mistake: Substituting without brackets: 3x² at x = −2 computed as 3 × −2² = 3 × 4 = 12? No — 3(−2)² = 12 but −2² without brackets reads −4.
The calculator evaluates −2² as −(2²).
Instead: Bracket every negative substitution: 3(−2)² − 2(−2) + 1 = 12 + 4 + 1 = 17.
Mistake: Writing an nth-term rule from one term only, e.g. 5, 8, 11 → 5n.
One term cannot fix a rule; the growth rate and the offset both matter.
Instead: Use the common difference for the coefficient of n, then adjust the constant to match the first term: 3n + 2.

Check yourself

Quick quiz

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Question 1 of 3

Question 1: current question
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Question 3: not answered

Expand and collect

−2(3x − 5) + 4x simplifies to

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5 multiple-choice questions and structured questions with answer lines. Chapter 4 (N5 Algebraic Expressions and Formulae: expansion, factorisation, algebraic fractions, nth term, formulae).

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Frequently asked questions

What does Sec 3 E-Math Chapter 4 cover?

It covers syllabus sub-topic N5 of O-Level 4052 / SEC G3 K310 in full: simplifying and expanding expressions, the three special identities, factorising linear and quadratic expressions, algebraic fractions with linear and quadratic denominators, nth-term patterns, and changing the subject of formulae.

How do I factorise 2x² + 7x + 3 without trial and error forever?

Multiply a by c (2 × 3 = 6) and find two numbers with that product and the middle-term sum: 6 and 1. Split the middle term (6x + x), factorise each half, and pull out the shared bracket: (x + 3)(2x + 1). Always finish by expanding to check.

Why do algebraic fractions need factorised denominators?

Because cancelling and common denominators work on whole factors. With x² − 9 written as (x − 3)(x + 3), you can see the shared factor with x − 3 and build the LCM in one glance. Unfactorised denominators hide all of that structure.

When is this chapter usually tested in school?

Algebra sits at the start of Term 2 in most schemes, so it typically lands in Weighted Assessment 2 and reappears in every later exam. Expansion, factorisation and fractions are also the machinery every later E-Math topic quietly assumes.

Sec 3 E-Math Algebraic Expressions & Formulae notes