What is the complete factorised form of (12a^2-27b^2)?

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

The direct answer is A: $3(2a-3b)(2a+3b)$. First take out the greatest common numerical factor, 3: $12a^2-27b^2=3(4a^2-9b^2)$. The expression inside is still factorable because $4a^2=(2a)^2$ and $9b^2=(3b)^2$. Apply $U^2-V^2=(U-V)(U+V)$: $4a^2-9b^2=(2a-3b)(2a+3b)$. Hence the complete form is $3(2a-3b)(2a+3b)$. Option C is only a partial factorisation; it leaves a difference of squares unfinished. Expanding A gives $3[4a^2-9b^2]=12a^2-27b^2$. B expands to $12a^2+12ab-27ab-27b^2$, not the original. D expands to $12a^2-9ab-18b^2$, also incorrect. The word complete means continue factoring until no common factor or standard identity remains.