What is the complete factorisation of (3a^2-12b^2)?

Correct answer: 3(a-2b)(a+2b)

Direct answer: C, \(3(a-2b)(a+2b)\). First take the common factor 3 from both terms: \(3a^2-12b^2=3(a^2-4b^2)\). Now recognise a difference of squares: \(a^2-4b^2=a^2-(2b)^2\). Using \(A^2-B^2=(A-B)(A+B)\), this becomes \((a-2b)(a+2b)\). Therefore the complete factorisation is \(3(a-2b)(a+2b)\). Option A stops after taking out 3 and is incomplete. Option B uses 4b, which would produce \(a^2-16b^2\) and hence \(3a^2-48b^2\). Option D expands to \(3a^2-9ab-12b^2\), introducing an unwanted ab term. Complete factorisation means continue until no common factor or applicable identity remains. A useful check is to multiply the two binomials first: their middle terms cancel and their last terms give \(-4b^2\).