Cambridge IGCSE Mathematics 0580: Introduction to algebra and algebraic manipulation – Study Guide

This study guide covers sections 2.1 Introduction to algebra and 2.2 Algebraic manipulation of the Cambridge IGCSE Mathematics 0580 syllabus for examination in 2025, 2026 and 2027. Sections C2.1/E2.1 and outcomes 1–3 of C2.2/E2.2 are the same for Core and Extended. Outcomes 4 and 5 of E2.2 (factorising five special forms, and completing the square) are Extended only. Core candidates take Paper 1 (non-calculator) and Paper 3 (calculator); Extended candidates take Paper 2 (non-calculator) and Paper 4 (calculator). This algebra is tested on both papers of your tier, including the non-calculator one, so you need to do it by hand.

For the whole of Topic 2, see the Algebra and graphs study guide and the Algebra and graphs revision notes. The course hub is at Cambridge IGCSE Mathematics, and you can tick off each outcome on the printable 0580 checklist. To find your gaps first, try the free Core diagnostic or Extended diagnostic.

What this unit covers

Syllabus	What you must be able to do	Tier
C2.1/E2.1 (1)	Know that letters can be used to represent generalised numbers	Core and Extended
C2.1/E2.1 (2)	Substitute numbers into expressions and formulas	Core and Extended
C2.2/E2.2 (1)	Simplify expressions by collecting like terms	Core and Extended
C2.2/E2.2 (2)	Expand products of algebraic expressions (Core: single brackets such as 3x(2x − 4y), and two brackets in one variable; Extended adds two brackets in two variables, and more than two brackets)	Core and Extended (wider on Extended)
C2.2/E2.2 (3)	Factorise by extracting common factors – “factorise” means factorise fully	Core and Extended
E2.2 (4)	Factorise ax + bx + kay + kby, a²x² − b²y², a² + 2ab + b², ax² + bx + c and ax³ + bx² + cx	Extended only
E2.2 (5)	Complete the square for expressions in the form ax² + bx + c	Extended only

The syllabus says that “simplify” means give the answer in its simplest form. The same rule applies across every paper: answers are expected in their simplest form unless the question says otherwise.

2.1 Letters as generalised numbers

A letter in algebra stands for a number. It is “generalised” because the statement is true whatever value the letter takes. When you write n + (n + 1) + (n + 2) for three consecutive whole numbers, you are describing every possible set of three consecutive numbers at once.

Vocabulary used in questions:

• A term is a number, a letter, or a product of them, such as 7, x, 5ab or −3x².
• An expression is one or more terms joined by + or −, such as 4x² − 3xy + 7. It has no equals sign.
• A formula links two or more quantities, such as A = lw. You can substitute into it.
• The coefficient is the number in front of the letters: in −3x², the coefficient is −3.

Shorthand: 3 × a is 3a, a × b is ab, a × a is a², and a ÷ b is a/b. Numbers go first: 5x, not x5.

Worked example 1: using a generalised number

Show that the sum of any three consecutive whole numbers is a multiple of 3.

Let the smallest number be n.
The three numbers are n, n + 1, n + 2.
Sum = n + n + 1 + n + 2
    = 3n + 3
    = 3(n + 1)
3(n + 1) is 3 times a whole number, so the sum is always a multiple of 3.

Testing 4 + 5 + 6 = 15 shows it works once; the algebra shows it works every time.

2.1 Substituting into expressions and formulas

To substitute, replace each letter with its value, then follow the order of operations (brackets, powers, multiplication and division, then addition and subtraction).

Two habits stop most errors:

1. Put negative numbers in brackets when you substitute.
2. Square before you multiply. In 3a², only a is squared, not 3a.

Watch the difference between (−3)² and −3². With the bracket, (−3)² = (−3) × (−3) = 9. Without it, −3² means −(3²) = −9. If a = −3, then a² = 9.

Worked example 2: substitution by hand (non-calculator)

Find the value of 3a² − 2ab when a = −2 and b = 5.

3a^2 - 2ab = 3(-2)^2 - 2(-2)(5)
           = 3(4) - (-20)
           = 12 + 20
           = 32

Taking away a negative is the same as adding.

Worked example 3: substitution with a calculator

The time, T seconds, for one swing of a pendulum of length l metres is given by T = 2π√(l/g). Find T when l = 0.8 and g = 9.8.

T = 2 x pi x sqrt(0.8 / 9.8)
  = 2 x pi x 0.28571...
  = 1.7951...
  = 1.80 (3 s.f.)

On Papers 3 and 4, non-exact answers must be given to 3 significant figures unless the question says otherwise. Keep the full value on your calculator until the end, and keep the trailing zero: 1.80, not 1.8. Use the π key or 3.142, as the syllabus allows.

2.2 Simplifying by collecting like terms

Like terms have exactly the same letters raised to exactly the same powers. 5xy and −3xy are like terms. x² and x are not, and neither are xy and x²y. You add or subtract the coefficients of like terms and leave the letters alone.

Worked example 4

Simplify 4x² − 3xy + 7 − x² + 5xy − 9.

x^2 terms:    4x^2 - x^2  = 3x^2
xy terms:    -3xy + 5xy   = 2xy
numbers:      7 - 9       = -2
Answer: 3x^2 + 2xy - 2

Each term carries the sign in front of it.

2.2 Expanding brackets

To expand, multiply every term inside the bracket by the term outside. For two brackets, multiply every term in the first by every term in the second, then collect like terms.

Worked example 5: two single brackets

Expand and simplify 5(2x − 3) − 2(x − 4).

5(2x - 3)  = 10x - 15
-2(x - 4)  = -2x + 8      (-2 times -4 is +8)
Total      = 10x - 15 - 2x + 8
           = 8x - 7

Worked example 6: two brackets in one variable

Expand and simplify (3x − 2)(2x + 5).

3x(2x + 5) = 6x^2 + 15x
-2(2x + 5) = -4x - 10
Total      = 6x^2 + 15x - 4x - 10
           = 6x^2 + 11x - 10

A squared bracket is two brackets: (2x − 3)² = (2x − 3)(2x − 3) = 4x² − 6x − 6x + 9 = 4x² − 12x + 9. Writing it as 4x² + 9 loses the middle term.

Worked example 7: two variables (Extended)

Expand and simplify (2x − y)(x + 3y).

2x(x + 3y) = 2x^2 + 6xy
-y(x + 3y) = -xy - 3y^2
Total      = 2x^2 + 5xy - 3y^2

Worked example 8: three brackets (Extended)

Expand and simplify (x + 1)(x − 3)(2x − 1).

Step 1: (x + 1)(x - 3) = x^2 - 3x + x - 3 = x^2 - 2x - 3
Step 2: (x^2 - 2x - 3)(2x - 1)
        x^2(2x - 1)  = 2x^3 - x^2
        -2x(2x - 1)  = -4x^2 + 2x
        -3(2x - 1)   = -6x + 3
Total = 2x^3 - 5x^2 - 4x + 3

Expand two brackets first, simplify, then multiply by the third. A three-term by two-term product gives six terms before collecting – count them to check none is missing.

2.2 Factorising by extracting common factors

Factorising is the reverse of expanding. Find the highest common factor (HCF) of the numbers and of each letter, write it outside a bracket, and divide each term by it to fill the bracket.

Worked example 9

Factorise fully 8x²y + 12xy² − 4xy.

HCF of 8, 12, 4 = 4
Each term has at least one x and one y, so take out xy
HCF = 4xy
8x^2y / 4xy = 2x,   12xy^2 / 4xy = 3y,   -4xy / 4xy = -1
Answer: 4xy(2x + 3y - 1)

Expand your answer to check it, and look inside the bracket for anything still in common. 2x(4xy + 6y² − 2y) is a correct factorisation, but not a full one, and earns only part of the marks. When a whole term is the common factor, as −4xy is here, a 1 (or −1) stays in the bracket.

Extended only: factorising the five special forms (E2.2, outcome 4)

Always take out any common factor first, then look for one of these structures.

Grouping: ax + bx + kay + kby

Pair the terms, factorise each pair, then take out the bracket they share.

Factorise 3px + 2qx + 6py + 4qy
= x(3p + 2q) + 2y(3p + 2q)
= (3p + 2q)(x + 2y)

If the brackets do not match, try a different pairing.

Difference of two squares: a²x² − b²y²

The pattern is a²x² − b²y² = (ax + by)(ax − by). Both terms must be perfect squares and they must be subtracted.

25m^2 - 16n^2 = (5m + 4n)(5m - 4n)

18x^2 - 50y^2 = 2(9x^2 - 25y^2)     common factor first
              = 2(3x + 5y)(3x - 5y)

A sum of two squares such as x² + 9 does not factorise this way.

Perfect square: a² + 2ab + b²

If the first and last terms are squares and the middle term is twice their product, the expression is a perfect square: a² + 2ab + b² = (a + b)².

9x^2 + 24x + 16:  9x^2 = (3x)^2,  16 = 4^2,  2(3x)(4) = 24x
= (3x + 4)^2

Quadratic: ax² + bx + c

When a = 1, find two numbers that multiply to c and add to b.

x^2 - 2x - 24:  numbers multiplying to -24 and adding to -2 are -6 and 4
= (x - 6)(x + 4)

When a ≠ 1, find two numbers that multiply to ac and add to b, split the middle term, then group.

6x^2 + x - 12:  ac = 6 x (-12) = -72
Numbers with product -72 and sum 1: 9 and -8
6x^2 + 9x - 8x - 12
= 3x(2x + 3) - 4(2x + 3)
= (3x - 4)(2x + 3)

Cubic with no constant: ax³ + bx² + cx

Every term contains x, so take out x first, then factorise the quadratic left behind.

2x^3 - 7x^2 - 15x = x(2x^2 - 7x - 15)
                  = x(2x + 3)(x - 5)

Check: (2x + 3)(x − 5) = 2x² − 10x + 3x − 15 = 2x² − 7x − 15.

Extended only: completing the square (E2.2, outcome 5)

Completing the square writes ax² + bx + c in the form a(x + p)² + q. It works because (x + p)² = x² + 2px + p², so x² + 2px = (x + p)² − p².

Worked example 10: a = 1

Write x² + 8x − 5 in the form (x + p)² + q.

Half of 8 is 4, so x^2 + 8x = (x + 4)^2 - 16
x^2 + 8x - 5 = (x + 4)^2 - 16 - 5
             = (x + 4)^2 - 21

Worked example 11: a ≠ 1

Write 2x² − 12x + 7 in the form a(x + p)² + q.

Take 2 out of the first two terms only:
2x^2 - 12x + 7 = 2(x^2 - 6x) + 7
x^2 - 6x = (x - 3)^2 - 9
         = 2[(x - 3)^2 - 9] + 7
         = 2(x - 3)^2 - 18 + 7
         = 2(x - 3)^2 - 11

The −9 sits inside the square bracket, so it is multiplied by 2 as well. Leaving it as −9 is the commonest slip. The same method handles a negative a: −x² + 6x + 1 = −(x² − 6x) + 1 = −[(x − 3)² − 9] + 1 = −(x − 3)² + 10. Expand your answer to check it.

Completed square form is used again to solve quadratic equations (E2.5).

Common errors

• Substituting a = −3 into a² and writing −9. Use brackets: (−3)² = 9.
• Collecting unlike terms, such as 3x² + 2x = 5x³.
• Losing the sign on the second bracket: −2(x − 4) is −2x + 8.
• Expanding (2x − 3)² as 4x² + 9 or 4x² − 9.
• (Extended) Taking out a numerical common factor and then missing the difference of two squares left inside.
• On Papers 3 and 4, rounding a substituted value early, or dropping a trailing zero from a 3 s.f. answer.

Next steps

Test yourself with the Algebra and graphs practice questions, then the Extended algebra practice if you are taking Papers 2 and 4. Use the Algebra and graphs revision notes for quick recall, and the Core diagnostic or Extended diagnostic to see which other resources for syllabus 0580 to work on next.

Official syllabus

Cambridge IGCSE Mathematics 0580 syllabus for examination in 2025, 2026 and 2027 (Version 3), published by Cambridge International – Topic 2 Algebra and graphs, sections C2.1/E2.1 Introduction to algebra and C2.2/E2.2 Algebraic manipulation.