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

This guide follows the OCR GCSE (9-1) Mathematics (J560) specification, Version 2.1 (August 2026). It covers content statements 6.01e (factorising), 6.01f (completing the square) and 6.01g (algebraic fractions). Common factors, quadratics of the form 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. Every skill here is done by hand.

Related pages: revision notes, practice questions, the OCR GCSE Mathematics hub, the printable checklist and the free 10-minute diagnostics.

What this unit covers

Content	What you must be able to do	Tier
6.01e	Take out common factors	Both tiers
6.01e	Factorise x² + bx + c, including the difference of two squares	Both tiers
6.01e	Factorise ax² + bx + c (a not 0 or 1)	Higher tier only
6.01f	Complete the square on a quadratic expression	Higher tier only
6.01g	Simplify and manipulate algebraic fractions	Higher tier only

Factorising is expanding brackets in reverse. You can always check an answer by expanding it. Expanding brackets itself (6.01d) belongs to the preceding unit.

Choosing a method for factorising (6.01e)

Ask these questions in order.

1. Is there a common factor? Take it out first.
2. How many terms are left? Two terms with a minus between perfect squares: difference of two squares. Three terms: a quadratic.
3. For a quadratic, is the coefficient of x² equal to 1? Use the multiply-and-add method. If not (Higher tier only), split the middle term.
4. Expand your answer to check it.

Common factors (6.01e)

Find the highest common factor of the numbers, then of each letter, and write it outside a bracket. What is left goes inside.

Example 1. Factorise fully 12x²y − 18xy².

Numbers: HCF of 12 and 18 is 6
x: lowest power is x
y: lowest power is y
12x²y − 18xy² = 6xy(2x − 3y)

Check: 6xy × 2x = 12x²y and 6xy × (−3y) = −18xy². Correct.

Example 2. Factorise 20a − 8a². The HCF is 4a. Then 20a ÷ 4a = 5 and 8a² ÷ 4a = 2a, so the answer is 4a(5 − 2a).

“Fully” means no common factor remains. 2(10a − 4a²) is not fully factorised.

Quadratics of the form x² + bx + c (6.01e)

Look for a pair of numbers whose product is c and whose sum is b. Then x² + bx + c = (x + m)(x + n).

c is	b is	The two numbers
positive	positive	two positive numbers
positive	negative	two negative numbers
negative	any	one of each sign. The larger (in size) takes the sign of b

Example 3. Factorise x² + 10x + 21. Pairs that multiply to 21: 1 and 21, 3 and 7. Only 3 + 7 = 10. Answer: (x + 3)(x + 7).

Example 4. Factorise x² − x − 42. Multiply to −42 and add to −1. Try 6 and −7: 6 × (−7) = −42 and 6 + (−7) = −1. Answer: (x − 7)(x + 6).

Example 5. Factorise x² − 12x + 32. Multiply to 32, add to −12, so both are negative: −4 and −8. Answer: (x − 4)(x − 8).

Difference of two squares (6.01e)

Two perfect squares with a minus between them factorise as

a² − b² = (a − b)(a + b)

There is no middle term because the two cross-terms cancel.

Example 6. x² − 49 = (x − 7)(x + 7).

Example 7. 9y² − 25. Here 9y² = (3y)² and 25 = 5², so the answer is (3y − 5)(3y + 5).

Example 8. 4x² − 100. Take out the common factor first: 4(x² − 25) = 4(x − 5)(x + 5).

Example 9. x² − 7. Seven is not a perfect square, but 7 = (√7)². So x² − 7 = (x − √7)(x + √7). The factors contain surds. This is the same idea as the specification’s example x² − 3.

A sum of squares such as x² + 49 does not factorise this way.

Factorising ax² + bx + c (6.01e, Higher tier only)

When a is not 0 or 1, split the middle term.

1. Multiply a by c.
2. Find two numbers that multiply to ac and add to b.
3. Rewrite bx as the sum of two x-terms using those numbers.
4. Factorise in two pairs. The two brackets must match.

Example 10. Factorise 3x² + 11x + 6.

ac = 3 × 6 = 18. Numbers that multiply to 18 and add to 11: 9 and 2
3x² + 9x + 2x + 6
= 3x(x + 3) + 2(x + 3)
= (3x + 2)(x + 3)

Example 11. Factorise 6x² − 7x − 5.

ac = 6 × (−5) = −30. Multiply to −30, add to −7: −10 and 3
6x² − 10x + 3x − 5
= 2x(3x − 5) + 1(3x − 5)
= (2x + 1)(3x − 5)

Write the “1” in the second pair. The bracket 3x − 5 is the whole second factor, so nothing is lost.

Example 12. Factorise fully 4x² + 10x − 6. Take out 2 first: 2(2x² + 5x − 3). Then ac = −6, and 6 and −1 give a sum of 5. So 2x² + 6x − x − 3 = 2x(x + 3) − (x + 3) = (2x − 1)(x + 3). Answer: 2(2x − 1)(x + 3).

Always check by expanding at least the first and last terms and the middle term.

Completing the square (6.01f, Higher tier only)

Completing the square turns x² + bx + c into the form (x + p)² + q.

x² + bx + c = (x + b/2)² + c − (b/2)²

Halve the coefficient of x to get p. Subtract p² to cancel the extra square. Then add the original constant.

Example 13. Write x² − 14x + 5 in the form (x + p)² + q.

Half of −14 is −7
(x − 7)² = x² − 14x + 49
x² − 14x + 5 = (x − 7)² − 49 + 5 = (x − 7)² − 44

Example 14. Write x² + 9x − 5 in the form (x + p)² + q.

Half of 9 is 9/2
x² + 9x − 5 = (x + 9/2)² − 81/4 − 5 = (x + 9/2)² − 101/4

When the coefficient of x² is not 1, take it out of the x² and x terms first. Leave the constant outside.

Example 15. Write 3x² + 12x + 5 in the form a(x + p)² + q.

3x² + 12x + 5 = 3(x² + 4x) + 5
              = 3[(x + 2)² − 4] + 5
              = 3(x + 2)² − 12 + 5
              = 3(x + 2)² − 7

Example 16. Write 2x² − 6x + 1 in the form a(x + p)² + q.

2x² − 6x + 1 = 2(x² − 3x) + 1
             = 2[(x − 3/2)² − 9/4] + 1
             = 2(x − 3/2)² − 9/2 + 1
             = 2(x − 3/2)² − 7/2

The fraction is multiplied by 2 as well. Forgetting this is the commonest slip. Check by expanding: 2(x − 3/2)² = 2x² − 6x + 9/2, and 9/2 − 7/2 = 1.

Algebraic fractions (6.01g, Higher tier only)

The rules are the same as for number fractions. Factorise first, because you can only cancel factors, never separate terms.

Simplifying

Example 17. Simplify (x² − 9)/(x² + x − 12).

Numerator:   x² − 9 = (x − 3)(x + 3)
Denominator: x² + x − 12 = (x + 4)(x − 3)
(x − 3)(x + 3) / [(x + 4)(x − 3)] = (x + 3)/(x + 4)

Example 18. Simplify (2x² + 7x + 3)/(4x² − 1). The numerator is (2x + 1)(x + 3). The denominator is (2x − 1)(2x + 1). Cancel (2x + 1) to get (x + 3)/(2x − 1).

Multiplying and dividing

To divide, turn the second fraction upside down and multiply. Factorise everything, then cancel.

Example 19. Simplify (x² + 5x)/(x² − 4) ÷ (x + 5)/(x + 2).

= x(x + 5)/[(x − 2)(x + 2)] × (x + 2)/(x + 5)
= x/(x − 2)

Adding and subtracting

Use a common denominator, usually the product of the two denominators. Write the answer as one fraction.

Example 20. Write 2/(x + 3) + 5/(x − 1) as a single fraction.

= [2(x − 1) + 5(x + 3)] / [(x + 3)(x − 1)]
= (2x − 2 + 5x + 15) / [(x + 3)(x − 1)]
= (7x + 13) / [(x + 3)(x − 1)]

Example 21. Write 3/(x − 2) − 2/(x + 1) as a single fraction. The numerator is 3(x + 1) − 2(x − 2) = 3x + 3 − 2x + 4 = x + 7. So the answer is (x + 7)/[(x − 2)(x + 1)]. The minus sign must act on the whole of 2(x − 2).

Example 22. Write 1/(x² − 4) + 3/(x + 2) as a single fraction. Since x² − 4 = (x − 2)(x + 2), the common denominator is (x − 2)(x + 2), not a longer product. The second fraction becomes 3(x − 2)/[(x + 2)(x − 2)]. The numerator is 1 + 3x − 6 = 3x − 5. Answer: (3x − 5)/[(x − 2)(x + 2)].

Common errors

• Cancelling terms, not factors. (x + 3)/(x + 4) cannot be reduced, and neither can (x² + 5)/x² to 5.
• Trying to factorise x² + 49 as a difference of two squares. The plus sign means it does not work.
• Leaving a common factor outside a fully factorised answer, such as 2(2x² + 5x − 3).
• Using the wrong sign pair: for c negative, the two numbers have different signs.
• Completing the square with 2x² + 8x: halving the 8 instead of halving 4 after taking out the 2.
• Not multiplying the subtracted fraction p² by the outside factor a.
• Losing a minus sign when subtracting fractions.

Next steps

Try the practice questions, then use the revision notes for quick recall. Tick off progress on the checklist.

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.