OCR GCSE Mathematics: Factorising, completing the square and algebraic fractions (J560) – Revision Notes

These notes condense OCR GCSE (9-1) Mathematics (J560) content 6.01e, 6.01f and 6.01g, from the Version 2.1 (August 2026) specification. For full explanations and longer worked examples, read the study guide. Test yourself with the practice questions. See also the OCR GCSE Mathematics hub, the checklist and the free diagnostics.

Common factors, x² + bx + c and the difference of two squares are on both tiers. Factorising ax² + bx + c, completing the square and algebraic fractions are Higher tier only.

Key results

Content	Result	Tier
6.01e common factor	ka + kb = k(a + b)	Both
6.01e quadratic	x² + bx + c = (x + m)(x + n) where mn = c and m + n = b	Both
6.01e difference of two squares	a² − b² = (a − b)(a + b)	Both
6.01e quadratic, a not 0 or 1	split bx using two numbers with product ac and sum b	Higher tier only
6.01f	x² + bx + c = (x + b/2)² + c − b²/4	Higher tier only
6.01g	factorise, then cancel common factors; common denominator to add or subtract	Higher tier only

Method boxes

Factorising: order of attack

1. Take out any common factor
2. Two terms, minus sign, both squares?  -> (a − b)(a + b)
3. Three terms, x² coefficient 1?        -> two numbers: product c, sum b
4. Three terms, x² coefficient not 1?    -> split the middle term (Higher tier only)
5. Expand to check

Sign guide for x² + bx + c

c	b	Numbers
+	+	two positives
+	−	two negatives
−	+	different signs, larger is +
−	−	different signs, larger is −

Splitting the middle term (Higher tier only)

8x² + 10x − 3
ac = −24. Product −24, sum +10: 12 and −2
8x² + 12x − 2x − 3
= 4x(2x + 3) − 1(2x + 3)
= (4x − 1)(2x + 3)

The brackets from the two pairs must be identical. If they are not, check the pair of numbers or the signs.

Completing the square (Higher tier only)

1. Take out a from the x² and x terms only (if a is not 1)
2. Halve the x coefficient: this is p
3. Write a(x + p)² and subtract a × p²
4. Add the original constant and tidy up

Worked reminder: x² − 8x + 3.

Half of −8 is −4
x² − 8x + 3 = (x − 4)² − 16 + 3 = (x − 4)² − 13

Try 5x² + 20x + 1 yourself. The answer is in the self-test.

Algebraic fractions (Higher tier only)

Simplify:     factorise top and bottom, cancel common FACTORS
Multiply:     factorise, multiply tops, multiply bottoms, cancel
Divide:       flip the second fraction, then multiply
Add/subtract: common denominator, combine numerators, simplify

Worked reminder: simplify (x² + x − 12)/(x² − 16).

Numerator:   (x + 4)(x − 3)
Denominator: (x − 4)(x + 4)
Cancel (x + 4):  (x − 3)/(x − 4)

Worked reminder: 1/x + 1/(x + 4).

= [(x + 4) + x] / [x(x + 4)] = (2x + 4)/[x(x + 4)]

You may leave the denominator factorised. Do not expand it unless asked.

Must-know distinctions

Idea	Take care
Factor and term	You cancel factors (things multiplied) never terms (things added)
x² − 25 and x² + 25	The first factorises. The second does not
Expand and factorise	They undo each other. Use one to check the other
4x² − 9 and 2x² − 18	Perfect squares 2x and 3 in the first. The second needs a common factor 2 first
(x + 3)² − 4 and x² + 6x + 5	These are equal. Completing the square changes the form, not the value
Simplify and write as a single fraction	Simplify means cancel. Single fraction means combine

Quick worked checks

Common factors. 15xy² + 10x²y has HCF 5xy. Answer: 5xy(3y + 2x).

Difference of two squares with surds. x² − 5 = (x − √5)(x + √5).

Completing the square with a fraction. x² + 5x + 4 = (x + 5/2)² − 25/4 + 4 = (x + 5/2)² − 9/4.

Check a completed square. Expand (x + 5/2)² − 9/4: x² + 5x + 25/4 − 9/4 = x² + 5x + 4. Correct.

Command words and what they ask for

Wording	What to give
Factorise	Brackets with no common factor left over, even if the question does not say “fully”
Factorise fully	The same. Check every bracket for a hidden common factor
Simplify (algebraic fraction)	One fraction with no common factor in top and bottom
Write as a single fraction	One fraction. Cancelling afterwards is a good habit
Write in the form (x + p)² + q	Find the numbers p and q
Write in the form a(x + p)² + q	Find the numbers a, p and q. Here a is the original x² coefficient

Number pairs worth knowing

Product	Pairs (positive)
12	1 and 12, 2 and 6, 3 and 4
16	1 and 16, 2 and 8, 4 and 4
18	1 and 18, 2 and 9, 3 and 6
24	1 and 24, 2 and 12, 3 and 8, 4 and 6
36	1 and 36, 2 and 18, 3 and 12, 4 and 9, 6 and 6

For a negative product, use the same pairs and give one of them a minus sign. Then pick the pair whose sum is b.

Squares to recognise: the squares of 2 up to 12, that is 4 up to 144. In algebra, x², 4x², 9x² and 25x² are squares of x, 2x, 3x and 5x.

Links between the skills

• Completing the square gives a check on a quadratic. When you expand (x + p)² + q, the middle term is 2p and the constant is p² + q. Use this to check your values.
• Difference of two squares appears inside fractions. A denominator such as x² − 16 usually signals that a numerator or another denominator will cancel with (x − 4) or (x + 4).
• Common factors come first everywhere. In a fraction, take common factors out of the top and bottom before searching for anything else.
• A quadratic that will not factorise is not a mistake. For example x² + 2x + 5 has no pair of whole numbers with product 5 and sum 2. Completing the square still works: (x + 1)² + 4.

Extra worked reminders

Higher tier only: fraction with a common factor to remove first. Simplify (4x² − 16)/(8x + 16).

Numerator:   4(x² − 4) = 4(x − 2)(x + 2)
Denominator: 8(x + 2)
Cancel 4 and (x + 2):  (x − 2)/2

Higher tier only: subtracting when one denominator is a difference of two squares. Write 4/(x − 3) − 6/(x² − 9) as a single fraction.

Common denominator (x − 3)(x + 3)
4(x + 3) − 6 = 4x + 12 − 6 = 4x + 6
Answer: (4x + 6)/[(x − 3)(x + 3)] = 2(2x + 3)/[(x − 3)(x + 3)]

Both tiers: three steps in one question. Factorise fully 3x² − 27. Take out 3: 3(x² − 9). Difference of two squares: 3(x − 3)(x + 3).

Quick self-test

Answer by hand, then check below.

1. Factorise 9a − 12b.
2. Factorise fully 10x² + 4x.
3. Factorise x² + x − 30.
4. Factorise x² − 64.
5. Factorise 4m² − 1.
6. Factorise 3x² + 10x + 8. (Higher tier only)
7. Write x² − 6x + 1 in the form (x + p)² + q. (Higher tier only)
8. Write 2x² + 8x − 3 in the form a(x + p)² + q. (Higher tier only)
9. Simplify (x² + 3x)/(x² − 9). (Higher tier only)
10. Write 2/(x − 1) − 3/(x + 2) as a single fraction. (Higher tier only)
11. Write 5x² + 20x + 1 in the form a(x + p)² + q. (Higher tier only)
12. Factorise 5x² + x − 4. (Higher tier only)

Answers

1. 3(3a − 4b)
2. 2x(5x + 2)
3. (x + 6)(x − 5)
4. (x − 8)(x + 8)
5. (2m − 1)(2m + 1)
6. (3x + 4)(x + 2). Using ac = 24, numbers 6 and 4: 3x² + 6x + 4x + 8 = 3x(x + 2) + 4(x + 2).
7. (x − 3)² − 8
8. 2(x + 2)² − 11. Working: 2(x² + 4x) − 3 = 2[(x + 2)² − 4] − 3 = 2(x + 2)² − 11.
9. x/(x − 3). Working: x(x + 3) / [(x − 3)(x + 3)].
10. (7 − x)/[(x − 1)(x + 2)]. Working: numerator 2(x + 2) − 3(x − 1) = 2x + 4 − 3x + 3 = 7 − x.
11. 5(x + 2)² − 19. Working: 5(x² + 4x) + 1 = 5[(x + 2)² − 4] + 1 = 5(x + 2)² − 19.
12. (5x − 4)(x + 1). Using ac = −20, numbers 5 and −4: 5x² + 5x − 4x − 4 = 5x(x + 1) − 4(x + 1).

Where marks are usually lost

• Stopping early: a common factor still sits inside a bracket, so the answer is not fully factorised.
• Choosing a number pair that adds to the right value but multiplies to the wrong one. Always test both.
• Sign slips with a negative c. The pair 3 and −8 gives −5, not +5.
• Treating x² + 16 as a difference of two squares.
• Missing a common factor before using the difference of two squares, as in 2x² − 18.
• Halving the wrong number when completing the square. You halve the coefficient of x, after taking out a.
• Multiplying the subtracted p² by a in the form a(x + p)² + q. Forgetting leaves the wrong constant.
• Cancelling a term, such as the x in (x + 3)/x, rather than a factor.
• Subtracting fractions and changing only the first sign of the numerator, not both terms of the bracket.
• Showing no working. Visible working lets you and the reader follow your reasoning, so write the factorised numerator and denominator before cancelling.

Official syllabus

OCR, GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), for first assessment in 2017. Content statements 6.01e, 6.01f and 6.01g.