Factorising Harder Quadratics

On the Higher paper you will meet quadratic expressions where the coefficient of x² is not 1, such as 6x² + 11x − 10. These are sometimes called non-monic quadratics and they require a more systematic approach than simple inspection. The most reliable method taught at GCSE is the ac method (also called splitting the middle term). This page walks you through every step with fully worked examples so you can tackle these confidently in the exam.

What Is Factorising Harder Quadratics?

A harder quadratic has the form ax² + bx + c where a ≠ 1. Because the leading coefficient is not 1, you cannot simply look for two numbers that multiply to c and add to b. Instead, you use the product ac to guide your factorisation, then split the middle term and factorise by grouping.

Key Formulas

Step-by-Step Method

1. Write the quadratic in the form ax² + bx + c. Identify a, b, and c.
2. Calculate the product ac.
3. Find two numbers that multiply to ac and add to b. List factor pairs of ac systematically.
4. Rewrite the middle term (bx) as the sum of two terms using those numbers.
5. Group the four terms into two pairs and factorise each pair.
6. Take out the common bracket to complete the factorisation.
7. Expand to check your answer matches the original expression.

Worked Example 1 — Foundation Level

Question: Factorise 2x² + 7x + 3.

Working:

• a = 2, b = 7, c = 3. So ac = 2 × 3 = 6.
• Find two numbers that multiply to 6 and add to 7: 1 and 6.
• Split the middle term: 2x² + 1x + 6x + 3.
• Group: (2x² + x) + (6x + 3).
• Factorise each group: x(2x + 1) + 3(2x + 1).
• Common bracket: (2x + 1)(x + 3).
• Check: 2x² + 6x + x + 3 = 2x² + 7x + 3 ✓

Answer: (2x + 1)(x + 3)

Worked Example 2 — Higher Level

Question: Factorise 6x² − 7x − 20.

Working:

• a = 6, b = −7, c = −20. So ac = 6 × (−20) = −120.
• Find two numbers that multiply to −120 and add to −7: 8 and −15 (since 8 × (−15) = −120 and 8 + (−15) = −7).
• Split: 6x² + 8x − 15x − 20.
• Group: (6x² + 8x) + (−15x − 20).
• Factorise: 2x(3x + 4) − 5(3x + 4).
• Common bracket: (3x + 4)(2x − 5).
• Check: 6x² − 15x + 8x − 20 = 6x² − 7x − 20 ✓

Answer: (3x + 4)(2x − 5)

Worked Example 3 — Exam Style

Question: Solve 3x² + 10x − 8 = 0.

Working:

First factorise. a = 3, b = 10, c = −8. ac = −24. Two numbers that multiply to −24 and add to 10: 12 and −2.

3x² + 12x − 2x − 8 = 0 3x(x + 4) − 2(x + 4) = 0 (x + 4)(3x − 2) = 0

So x + 4 = 0 giving x = −4, or 3x − 2 = 0 giving x = 2/3.

Answer: x = −4 or x = 2/3

Common Mistakes

• Using the wrong product. Students sometimes look for numbers that multiply to c instead of ac. Always calculate ac first.
• Sign errors when splitting the middle term. Be very careful with negatives. If ac is negative, one of your two numbers must be positive and the other negative.
• Incorrect grouping. When factorising by grouping, both groups must produce the same bracket. If they do not, check your splitting step.

Exam Tips

• Write out factor pairs of ac methodically — do not guess. This avoids wasting time.
• If you cannot find a pair, double-check your values of a, b, and c, especially signs.
• The ac method always works for factorisable quadratics, so it is the most reliable approach.
• After factorising, the question may ask you to solve the equation — set each bracket equal to zero.

Practice Questions

Q1 (Foundation): Factorise 2x² + 5x + 2.

Q2 (Higher): Factorise 5x² − 13x − 6.

Q3 (Higher): Solve 4x² + 4x − 3 = 0.

Practise harder quadratics with instant feedback free on GCSEMathsAI.

Related Topics

• Factorising Expressions
• Factorising Quadratics
• Solving Quadratic Equations by Factorising

Summary

• Harder quadratics have the form ax² + bx + c where a is not 1.
• Use the ac method: find two numbers that multiply to ac and add to b.
• Split the middle term, group in pairs, and factorise each group.
• Always expand your answer to verify it matches the original expression.
• This skill is essential for solving equations, simplifying algebraic fractions, and completing the square.