Expanding Double Brackets

Grade 5

Multiplying two brackets

When you have two brackets multiplied together — like (x + 3)(x + 5) — every term in the first bracket must multiply every term in the second. With two terms in each bracket, that gives four multiplications.

The FOIL method is a handy order: Firsts, Outsides, Insides, Lasts. Then collect like terms.

Worked example — FOIL

EXAMPLE

Expand (x + 3)(x + 5).

Firsts: x × x = x². Outsides: x × 5 = 5x. Insides: 3 × x = 3x. Lasts: 3 × 5 = 15. Add them all: x² + 5x + 3x + 15 = x² + 8x + 15.

Difference of two squares — a special case

When the two brackets are (a + b)(a − b), the middle terms cancel. This gives the difference of two squares result: (a + b)(a − b) = a² − b². It is worth memorising — examiners love it.

EXAMPLE

Expand (x + 7)(x − 7).

Using difference of two squares: x² − 7² = x² − 49.

Watch out with negatives

In (x − 4)(x − 3), the Lasts multiplication is (−4)(−3) = +12, not −12. Keep track of signs at every step — this is where most marks are lost.

Try this one

(a) Factorise \( x^2 - 3x - 10 \). [2]

(b) Hence solve \( x^2 - 3x - 10 = 0 \). [1]

Nova's hint:

Part (a) uses factorising a quadratic — find two numbers that multiply to the constant term and add to the coefficient of \( x \).

Ask yourself: 'Once I have the factorised form \( (x + a)(x + b) = 0 \), what must be true about at least one of the brackets for the product to equal zero?'

Try to Solve it with Nova?