Expanding brackets and factorising — including quadratics

Expanding goes from a product to a sum; factorising goes from a sum back to a product. The two are inverse operations and you must be quick at both.

Expanding single brackets

Multiply every term inside by the term outside. 3(2x - 5) = 6x - 15. Negative outside flips signs: -2(x - 4) = -2x + 8.

Expanding double brackets — FOIL

(x + 3)(x + 7): First, Outer, Inner, Last. F: x × x = x². O: x × 7 = 7x. I: 3 × x = 3x. L: 3 × 7 = 21. Sum: x² + 7x + 3x + 21 = x² + 10x + 21.

For the special products, memorise:

• (x + a)² = x² + 2ax + a² — the middle term is double the product.
• (x - a)² = x² - 2ax + a² — same form, leading minus on the middle.
• (x + a)(x - a) = x² - a² — the difference of two squares.

Factorising — single bracket (taking out a common factor)

Look for the highest factor common to every term. 6x² - 9x = 3x(2x - 3). Always check by re-expanding.

Factorising quadratics with leading coefficient 1

To factorise x² + bx + c, find two numbers that multiply to c and add to b.

Example: x² + 7x + 12. Need product 12, sum 7. → 3 and 4. So (x + 3)(x + 4).

Example with negatives: x² - 5x + 6. Product 6, sum -5. → -2 and -3. So (x - 2)(x - 3).

Example with sign change: x² + x - 6. Product -6, sum +1. → +3 and -2. So (x + 3)(x - 2).

Difference of two squares

x² - 49 = (x + 7)(x - 7). Recognise it: a square minus a square. Don't try the long way.

Factorising quadratics with leading coefficient ≠ 1 [H]

2x² + 7x + 3. Multiply leading coefficient by constant: 2 × 3 = 6. Find two numbers multiplying to 6, summing to 7: 1 and 6. Split the middle term: 2x² + 1x + 6x + 3. Group: x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3).

Always check by re-expanding.