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

These notes condense OCR GCSE (9-1) Mathematics, specification Version 2.1 (August 2026), sections 6.01a to 6.01d. For full explanations and longer worked examples, read the study guide. Content marked Higher tier only is for Higher tier candidates alone; everything else is for both tiers.

Use the OCR GCSE Mathematics hub and the printable checklist to track what you have revised. When you are ready, try the practice questions and the free diagnostics.

Vocabulary (6.01a)

Word	Remember
Term	A part of an expression with its sign: in 2x² − 7x + 1 the terms are 2x², −7x and +1
Expression	No equals sign. Simplify it or expand it
Equation	Has =. True only for certain values. You solve it
Formula	Equation linking real quantities, such as v = u + at
Inequality	Uses <, >, ≤ or ≥
Identity	True for all values. Written ≡
Factor	Divides exactly into something. 4 and x are factors of 12xy

Equivalence and arguments (6.01a, both tiers)

Method to show two expressions are equivalent:

1. Pick the more complicated side.
2. Expand and simplify it, one step per line.
3. Stop when it matches the other side exactly.
4. Write a short closing sentence.

Reminder: show (x + 2)² − (x − 2)² ≡ 8x.

(x² + 4x + 4) − (x² − 4x + 4) = 8x

To disprove a claim, one counterexample is enough. For (x + 2)² = x² + 4, try x = 2: the left is 16 and the right is 8.

To construct an argument, name the quantities with letters, build the expression, simplify, and say what the result tells you.

Algebraic proof (6.01a, Higher tier only)

Idea	Algebra
Any even number	2n
Any odd number	2n + 1
Next even or odd number	add 2
Consecutive integers	n, n + 1, n + 2
Two different integers	two letters, such as a and b

Method in steps:

1. Define the numbers with letters that stand for integers.
2. Form the expression and expand.
3. Factorise to show the required factor.
4. Write a closing sentence: “this is 4 times an integer, so it is a multiple of 4”.

Worked reminder: the sum of two consecutive odd numbers is a multiple of 4.

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

Another: the square of an odd number is one more than a multiple of 4.

(2n + 1)² = 4n² + 4n + 1 = 4n(n + 1) + 1

Examples are never a proof. Testing 3, 5 and 7 shows nothing about all odd numbers.

Collecting like terms (6.01b)

Like terms have the same letters with the same powers. Keep the sign with its term.

Terms	Like?
4ab and −ba	Yes
3x and 3x²	No
2xy and 2x	No

Worked reminder: 6x + 2y − x + 4y = 5x + 6y.

Products and quotients (6.01c)

Both tiers

Rule	Example
Multiply numbers, then letters	2a × 3b = 6ab
aᵐ × aⁿ = aᵐ⁺ⁿ	a² × a³ = a⁵
aᵐ ÷ aⁿ = aᵐ⁻ⁿ	3a³ ÷ a² = 3a
(aᵐ)ⁿ = aᵐⁿ	(a²)⁴ = a⁸

Method in steps: numbers first, then each letter in turn, then write the answer in alphabetical order.

Worked reminder: 3a² × 4a³ = 12a⁵, and 20m⁵n ÷ 5m² = 4m³n.

Negative and fractional powers (Higher tier only)

Rule	Example
a⁻ⁿ = 1/aⁿ	x⁻² = 1/x²
a^(1/2) = √a	25^(1/2) = 5
a^(1/n) = nth root of a	27^(1/3) = 3
a^(m/n) = (a^(1/n))ᵐ	8^(2/3) = 4

Method in steps for a fractional power of a product: deal with the number and each letter separately. For a negative power, work out the positive power first and then take the reciprocal.

Worked reminder: 2a⁻³ × 3a⁵ = 6a², and (16x⁸)^(3/4) = 8x⁶, because 16^(3/4) = 8 and x⁸ to the power ¾ is x⁶.

Expanding brackets (6.01d)

Single bracket (both tiers)

Multiply every term inside by the term outside, with its sign. A minus sign in front of a bracket flips every sign inside.

Worked reminder: 5(x − 2) − (3 − x) = 5x − 10 − 3 + x = 6x − 13.

Two binomials (both tiers)

Multiply each term in the first bracket by each term in the second (four products), then collect like terms.

Worked reminder: (x + 5)(x − 3) = x² − 3x + 5x − 15 = x² + 2x − 15.

Special patterns, which you can always expand by the grid method:

Pattern	Result
(a + b)²	a² + 2ab + b²
(a − b)²	a² − 2ab + b²
(a + b)(a − b)	a² − b²

More than two binomials (Higher tier only)

Expand two brackets, tidy, then multiply by the third. Check by substituting x = 1 or x = 2 into the original and the answer.

Worked reminder: (x − 1)(x + 1)(x + 3).

(x − 1)(x + 1) = x² − 1
(x² − 1)(x + 3) = x³ + 3x² − x − 3

Check with x = 2: the original gives 1 × 3 × 5 = 15, and the answer gives 8 + 12 − 2 − 3 = 15.

What the command words ask for

Word	What to do
Simplify	Collect like terms and combine powers until nothing more can be joined
Expand	Remove the brackets by multiplying out
Expand and simplify	Remove the brackets, then collect like terms
Show that	Start from one side and reach the stated result, with every line visible
Prove (Higher tier only)	Use letters for general numbers and end with a sentence
Write in the form	Rewrite so that the answer matches the pattern given

Quick order of work for a mixed expression

Take 2x(3x − 1) + (x + 4)(x − 2) as an example.

2x(3x − 1)     = 6x² − 2x
(x + 4)(x − 2) = x² − 2x + 4x − 8 = x² + 2x − 8
Add:  6x² − 2x + x² + 2x − 8 = 7x² − 8

1. Expand each part separately.
2. Write each result in full before joining them.
3. Collect like terms at the end.
4. Substitute a small number such as x = 1 as a check: the original gives 2 × 2 + 5 × (−1) = −1, and the answer gives 7 − 8 = −1.

Must-know distinctions

Pair	Difference
Expression and equation	An expression is simplified; an equation is solved
Equation and identity	Some values compared with all values
Proof and checking	Proof uses letters and covers every case; checking uses numbers and does not
Adding and multiplying indices	Add indices when multiplying terms; multiply indices for a power of a power
x⁻² and −x²	1/x² compared with minus x squared
(a + b)² and a² + b²	Missing 2ab in the second
2a × 3b and 2a + 3b	6ab compared with 2a + 3b, which cannot be simplified
a^(1/2) and a²	A square root compared with a square
Term and factor	Terms are added; factors are multiplied

Quick self-test

1. Simplify 6x + 2y − x + 4y.
2. State whether 3(x + 1) = 3x + 3 is an equation or an identity, and say why.
3. Simplify 3a² × 4a³.
4. Simplify 20m⁵n ÷ 5m².
5. (Higher tier only) Simplify 2a⁻³ × 3a⁵.
6. (Higher tier only) Simplify (16x⁸)^(3/4).
7. Expand and simplify 5(x − 2) − (3 − x).
8. Expand (x + 5)(x − 3).
9. (Higher tier only) Expand (x − 1)(x + 1)(x + 3).
10. (Higher tier only) Prove that the sum of two consecutive odd numbers is a multiple of 4.
11. Show that (x + 2)² − (x − 2)² ≡ 8x.
12. Give a value of x that shows (x + 2)² = x² + 4 is not an identity.

Answers

1. 5x + 6y.
2. An identity, because it is true for every value of x (the left side expands to the right side).
3. 12a⁵.
4. 4m³n.
5. 6a².
6. 8x⁶.
7. 6x − 13.
8. x² + 2x − 15.
9. x³ + 3x² − x − 3.
10. Let the numbers be 2n + 1 and 2n + 3. The sum is 4n + 4 = 4(n + 1), which is 4 times an integer.
11. (x² + 4x + 4) − (x² − 4x + 4) = 8x.
12. x = 2 gives 16 on the left and 8 on the right. Any value other than 0 works.

Where marks are usually lost

• Not showing the expansion line in a “show that” or proof question, so there is nothing for the reader to credit.
• Using one letter for two different integers in a proof.
• Finishing a proof with a correct expression but no sentence saying why it proves the claim.
• Losing a minus sign when a bracket has a negative number or a minus sign in front.
• Writing (x + 3)² as x² + 9, or (x − 4)² as x² − 16.
• Adding indices that should be multiplied, or the other way round.
• Reading a negative power as a negative number.
• Stopping after expanding, without collecting like terms.
• Making a slip in the third bracket of a triple expansion because the first two were not tidied.
• Giving a few numerical checks as the whole proof.
• Writing the final answer in a different form from the one asked for, such as a negative power when a fraction was requested.

Official syllabus

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