Factorise (4x^2-9).

Correct answer: (2x-3)(2x+3)

The direct answer is A: \((2x-3)(2x+3)\). Recognise \(4x^2-9\) as a difference of two squares. The first term is \(4x^2=(2x)^2\) and the second is \(9=3^2\). Apply \(P^2-Q^2=(P-Q)(P+Q)\), with \(P=2x\) and \(Q=3\). Thus \(4x^2-9=(2x-3)(2x+3)\). Expanding confirms it: \(4x^2+6x-6x-9=4x^2-9\), because the middle terms cancel. B incorrectly uses 4x and 9 as the square roots and expands to \(16x^2-81\). C is a square, and its expansion is \(4x^2-12x+9\), which contains a middle term. D expands to \(4x^2-9x-9\), also not the given expression. Memory cue: for two squares with a minus sign, write the difference and the sum of their actual square roots.