OCR GCSE Mathematics: Algebraic language, proof, simplifying and expanding (J560)

This guide is for OCR GCSE (9-1) Mathematics, aligned to the OCR specification Version 2.1 (August 2026). It covers sections 6.01a to 6.01d: algebraic terminology and proofs, collecting like terms, simplifying products and quotients, and multiplying out brackets. Most of it is for both tiers. Content marked Higher tier only is for Higher tier candidates alone. These are algebra skills, so every example is done by hand.

Check your progress with the printable checklist and browse the OCR GCSE Mathematics hub. Then use the revision notes and the practice questions. Not sure where you stand? Try the free diagnostics. For the number side of indices see number operations and integers.

What this unit covers

Section	What you must be able to do	Tier
6.01a	Use the words expression, equation, formula, inequality, term and factor correctly	Both tiers
6.01a	Tell an equation from an identity; show two expressions are equivalent; construct algebraic arguments	Both tiers
6.01a	Use algebra to construct proofs	Higher tier only
6.01b	Simplify by collecting like terms	Both tiers
6.01c	Simplify products and quotients, including using index laws with positive whole-number powers	Both tiers
6.01c	Use the laws of indices with negative and fractional powers	Higher tier only
6.01d	Expand a single bracket; expand the product of two binomials	Both tiers
6.01d	Expand the product of more than two binomials	Higher tier only

Algebraic language (6.01a)

Precise words let you read a question correctly. Learn these.

Word	Meaning	Example
Term	One part of an expression, with its sign. Terms are separated by + and −	5x², −3x and 8 are the terms of 5x² − 3x + 8
Expression	Terms combined with no equals sign. You simplify or expand it, but you cannot solve it	4a + 7
Equation	Two expressions joined by =, true for particular values of the letter	3x + 5 = 17 is true only when x = 4
Formula	An equation linking quantities that have a meaning	A = ½bh
Inequality	Two expressions joined by <, >, ≤ or ≥	4t > 20
Factor	A number or expression that divides exactly into another, or that is multiplied to make it	3 and a are factors of 12ab

The number in front of a letter is its coefficient. In 5x² the coefficient is 5.

Equation or identity?

An equation is true only for some values. An identity is true for every value of the letters. We write an identity with the sign ≡, though = is often used in questions.

• 3(x + 2) ≡ 3x + 6 is an identity. Any value of x works.
• 5x − 1 = 14 is an equation. Only x = 3 works.

Showing that expressions are equivalent

Take one side and change it, line by line, until it matches the other. Do not work on both sides at once. Do not stop after trying one number.

Worked example 1. Show that (x − 2)² + 3 ≡ x² − 4x + 7.

(x − 2)² + 3 = (x − 2)(x − 2) + 3
             = x² − 2x − 2x + 4 + 3
             = x² − 4x + 7

The left side now equals the right side, so the two are equivalent.

A single test value can show two expressions are not equivalent, but it cannot show they are. If someone claims (x + 2)² = x² + 4, try x = 1: the left is 9 and the right is 5. One failure is enough to disprove it.

Constructing an algebraic argument

Write what is given using letters, simplify, then say what the result means.

Worked example 2. A rectangle has length 2x + 5 and width x − 1. Show that its perimeter is 6x + 8.

Perimeter = 2 × (length + width)
          = 2 × (2x + 5 + x − 1)
          = 2 × (3x + 4)
          = 6x + 8

Algebraic proof (Higher tier only)

A proof must hold for every number of its kind, so you cannot use examples. Use these standard forms, where n, k, a and b are integers:

Idea	Algebra
Even number	2n
Odd number	2n + 1
Consecutive integers	n, n + 1, n + 2, …
Two different integers	use two letters, such as a and b

Method: write the expression, simplify it, then show a common factor that proves the claim. End with a sentence.

Worked example 3. Prove that the sum of five consecutive integers is a multiple of 5.

n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 5n + 10
                                          = 5(n + 2)

n + 2 is an integer, so the sum is 5 times an integer. It is a multiple of 5. The same method with three consecutive integers gives 3n + 3 = 3(n + 1), a multiple of 3.

Worked example 4. Prove that the product of two odd numbers is odd.

Let the numbers be 2a + 1 and 2b + 1. Use different letters, because the numbers need not be equal.

(2a + 1)(2b + 1) = 4ab + 2a + 2b + 1
                 = 2(2ab + a + b) + 1

2ab + a + b is an integer, so this has the form 2 × integer + 1. It is odd.

Worked example 5. Prove that the difference between the squares of two consecutive odd numbers is a multiple of 8.

Let the numbers be 2k + 1 and 2k + 3.

(2k + 3)² − (2k + 1)² = 4k² + 12k + 9 − (4k² + 4k + 1)
                      = 8k + 8
                      = 8(k + 1)

k + 1 is an integer, so the difference is a multiple of 8.

Collecting like terms (6.01b)

Like terms have exactly the same letters with exactly the same powers. Add or subtract the coefficients and keep the letters. The sign in front of a term belongs to it.

• 3x and 8x are like terms. 3x and 3x² are not. 3x and 3y are not.
• xy and yx are like terms, because multiplication can be done in either order.

Worked example 6. Simplify 7a − 3b + 2a + 5b − a.

a-terms: 7a + 2a − a = 8a
b-terms: −3b + 5b = 2b

Answer: 8a + 2b.

Worked example 7. Simplify 4x² + 3x − 2 − x² − 5x + 6.

x² terms:   4x² − x² = 3x²
x terms:    3x − 5x = −2x
numbers:    −2 + 6 = 4

Answer: 3x² − 2x + 4.

Worked example 8. Simplify 5xy − 2yx + x².

5xy − 2yx = 3xy, and x² is a different kind of term. Answer: 3xy + x².

Simplifying products and quotients (6.01c)

Both tiers

To multiply terms, multiply the numbers, then multiply each letter, adding indices. To divide, divide the numbers, then subtract indices.

Rule	Example
a × a × a = a³	x × x = x²
aᵐ × aⁿ = aᵐ⁺ⁿ	a² × a³ = a⁵
aᵐ ÷ aⁿ = aᵐ⁻ⁿ	3a³ ÷ a² = 3a
(aᵐ)ⁿ = aᵐⁿ	(a³)² = a⁶

Worked example 9. Simplify 5x²y × 2xy³.

Numbers: 5 × 2 = 10. x: x² × x = x³. y: y × y³ = y⁴. Answer: 10x³y⁴.

Worked example 10. Simplify 18a⁴b³ ÷ 6a²b.

Numbers: 18 ÷ 6 = 3. a: a⁴ ÷ a² = a². b: b³ ÷ b = b². Answer: 3a²b².

Worked example 11. Simplify 12p³q / 8pq².

Divide the numbers by their common factor 4 to get 3/2. p: p³ ÷ p = p². q: q ÷ q² leaves q on the bottom. Answer: 3p² / 2q.

Negative and fractional powers (Higher tier only)

Rule	Meaning
a⁻ⁿ = 1/aⁿ	a negative power is a reciprocal
a^(1/2) = √a	a power of ½ is a square root
a^(1/n)	the nth root of a
a^(m/n)	take the root, then the power: (a^(1/n))ᵐ

The first three index laws still apply. Add indices when multiplying and subtract when dividing, even when they are negative or fractions.

Worked example 12. Simplify 3x⁵ × 4x⁻².

12 × x^(5 + (−2)) = 12x³.

Worked example 13. Write 8y² ÷ 2y⁵ with a negative power and as a fraction.

4 × y^(2 − 5) = 4y⁻³ = 4 / y³.

Worked example 14. Simplify (9x⁴)^(1/2).

The root of 9 is 3, and x⁴ to the power ½ is x². Answer: 3x².

Worked example 15. Simplify (8a⁶)^(−1/3).

The negative sign means reciprocal. The cube root of 8 is 2, and a⁶ to the power ⅓ is a². So the answer is 1 / 2a².

Worked example 16. Simplify 4a³b⁻² × 5a⁻⁵b^(1/2).

Numbers: 20. a: 3 + (−5) = −2. b: −2 + ½ = −3/2. Answer: 20a⁻²b^(−3/2).

Multiplying out brackets (6.01d)

Single brackets (both tiers)

Multiply every term inside by the term outside, including its sign.

Worked example 17. Expand 3(2a − 5) and −2(3x − 4).

3(2a − 5) = 6a − 15. For the second, −2 × 3x = −6x and −2 × (−4) = +8, giving −6x + 8.

Worked example 18. Expand and simplify 4(2x − 1) − 3(x − 5).

4(2x − 1) − 3(x − 5) = 8x − 4 − 3x + 15
                     = 5x + 11

The minus before the second bracket changes both signs inside it.

A term outside can be a letter: 3x(2x² − 5) = 6x³ − 15x.

Two binomials (both tiers)

A binomial has two terms. Multiply each term in the first bracket by each term in the second, then collect like terms. A grid keeps the four products organised.

Worked example 19. Expand (2x − 3)(x + 5).

×	x	+5
2x	2x²	+10x
−3	−3x	−15

Add: 2x² + 10x − 3x − 15 = 2x² + 7x − 15.

Worked example 20. Expand (3x − 2)².

Squaring means multiplying by itself, so write (3x − 2)(3x − 2) and never 9x² + 4.

9x² − 6x − 6x + 4 = 9x² − 12x + 4

Worked example 21. Expand (a + 2b)(a − 3b).

a² − 3ab + 2ab − 6b² = a² − ab − 6b².

More than two binomials (Higher tier only)

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

Worked example 22. Expand (x + 2)(x − 3)(x + 4).

(x + 2)(x − 3) = x² − x − 6
(x² − x − 6)(x + 4) = x³ + 4x² − x² − 4x − 6x − 24
                    = x³ + 3x² − 10x − 24

Cubing a bracket works the same way. (x − 2)³ = (x − 2)(x − 2)(x − 2). The first two give x² − 4x + 4, and then (x² − 4x + 4)(x − 2) = x³ − 6x² + 12x − 8.

A check: put x = 1 into the original and the answer. In worked example 22, the left gives 3 × (−2) × 5 = −30 and the right gives 1 + 3 − 10 − 24 = −30. If the values differ, you made an error.

Common errors

• Writing (x + 3)² = x² + 9. The middle term 6x is lost.
• Forgetting the sign of the outside term, so −2(3x − 4) becomes −6x − 8.
• Treating 3x and 3x² as like terms.
• Multiplying indices when you should add them: x² × x³ is x⁵, not x⁶.
• Writing 2a × 3b as 5ab. Multiply the numbers: 6ab.
• In a proof, using one letter for two different numbers, or checking a few examples and stopping.
• Ending a proof without saying why the final form shows the property.
• Treating a negative power as a negative number. x⁻² is 1/x², not −x².

Next steps

Condense this unit with the revision notes, then test yourself on the practice questions. Factorising and algebraic fractions follow in 6.01e to 6.01g.

Official syllabus

OCR GCSE (9-1) Mathematics (J560), Version 2.1 (August 2026), for first assessment in 2017, published by OCR. This guide covers sections 6.01a to 6.01d.