Some GCSE equations require expanding double brackets first. When both brackets contain x, expanding always produces an x² term — the result is a quadratic. Collect all terms on one side to form ax² + bx + c = 0, then solve by factorising or the quadratic formula. Recognising this two-stage structure is the key skill.
Why do double brackets lead to quadratic equations?
When you multiply (x + a)(x + b), the FOIL method (First, Outside, Inside, Last) gives:
- First: x × x = x²
- Outside: x × b = bx
- Inside: a × x = ax
- Last: a × b = ab
Result: x² + (a + b)x + ab
Because x appears in both brackets, multiplying them always generates an x² term. Any equation where this product equals a number or another expression will be quadratic.
How do you solve an equation with double brackets?
Follow these three stages:
- Expand the double brackets using FOIL.
- Rearrange to form ax² + bx + c = 0 (bring all terms to one side).
- Solve the quadratic by factorising, completing the square, or the quadratic formula.
Worked example 1: (x + 3)(x − 2) = 10
| Step | Working |
|---|---|
| Expand | x² − 2x + 3x − 6 = 10 |
| Simplify | x² + x − 6 = 10 |
| Rearrange to = 0 | x² + x − 16 = 0 |
| Discriminant check | b² − 4ac = 1 + 64 = 65 (not a perfect square → no integer factors) |
| Quadratic formula | x = (−1 ± √65) / 2 |
| Solutions | x = (−1 + 8.06…) / 2 ≈ 3.53 or x = (−1 − 8.06…) / 2 ≈ −4.53 |
Check: (3.53 + 3)(3.53 − 2) = (6.53)(1.53) ≈ 9.99 ≈ 10 ✓
Worked example 2: (2x − 1)(x + 4) = x + 3
| Step | Working |
|---|---|
| Expand | 2x² + 8x − x − 4 = x + 3 |
| Simplify left side | 2x² + 7x − 4 = x + 3 |
| Move all terms left | 2x² + 7x − x − 4 − 3 = 0 |
| Simplify | 2x² + 6x − 7 = 0 |
| Factorise if possible | Discriminant: 36 + 56 = 92 — not a perfect square |
| Quadratic formula | x = (−6 ± √92) / 4 |
| Solutions | x ≈ 0.65 or x ≈ −5.40 (to 2 d.p.) |
Worked example 3: a case that factorises neatly
Solve (x + 5)(x − 1) = 5.
| Step | Working |
|---|---|
| Expand | x² − x + 5x − 5 = 5 |
| Simplify | x² + 4x − 5 = 5 |
| Rearrange | x² + 4x − 10 = 0 |
| Discriminant | 16 + 40 = 56 — not a perfect square |
Hmm — let's try (x + 5)(x − 1) = −5 instead (a different question):
| Step | Working |
|---|---|
| Expand | x² + 4x − 5 = −5 |
| Rearrange | x² + 4x = 0 |
| Factorise | x(x + 4) = 0 |
| Solutions | x = 0 or x = −4 |
Check x = 0: (0 + 5)(0 − 1) = 5 × (−1) = −5 ✓
Which solving method should you use?
| Form of quadratic after rearranging | Method |
|---|---|
| Discriminant is a perfect square | Factorise |
| ax² + bx + c = 0, a = 1, factors are clear | Factorise |
| Asks for exact values | Completing the square or quadratic formula |
| Asks for decimal answers or graph sketching | Quadratic formula |
Frequently asked questions
How do I check whether the quadratic will factorise?
Calculate the discriminant b² − 4ac. If the result is a perfect square (0, 1, 4, 9, 16, 25, …), the quadratic factorises over the integers. If it is not a perfect square, the roots are irrational and you need the quadratic formula or completing the square.
What if there are three brackets, e.g. (x + 1)(x + 2)(x − 3)?
Expand two of the brackets first to get a quadratic, then multiply by the third bracket to get a cubic. The equation becomes cubic (degree 3) and is a Higher tier stretch problem. The standard approach is to try integer roots (by the factor theorem) and then factorise the remaining quadratic.
Why must I bring all terms to one side to form = 0?
The factorising and quadratic formula methods only apply when one side is zero. Leaving the equation as x² + x − 6 = 10 and factorising the left side to (x + 3)(x − 2) = 10 does not help — even though the left side factorises, you cannot set each factor equal to 10 and solve separately.
What is the most common error in these questions?
Expanding the brackets incorrectly — particularly missing the Outside or Inside terms in FOIL, or making sign errors with negative terms. Write out all four FOIL terms separately before collecting them. For (2x − 1)(x + 4), explicitly write: 2x², 8x, −x, −4 before simplifying to 2x² + 7x − 4.
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