Choose the factorised form of (x^2+4xy+4y^2-9z^2).

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

The direct answer is option A: \((x+2y−3z)(x+2y+3z)\). First recognise \(x^2+4xy+4y^2\) as \((x+2y)^2\), using the square identity. Also, \(9z^2=(3z)^2\). The expression therefore becomes \((x+2y)^2−(3z)^2\). Apply the difference-of-squares identity \(U^2−V^2=(U−V)(U+V)\), with \(U=x+2y\) and \(V=3z\). This gives \((x+2y−3z)(x+2y+3z)\), option A. Option B uses \(x−2y\), whose square would produce \(-4xy\), so it has the wrong middle sign. Option C is not a complete factorisation of the difference of two squares and also writes \(-3z\) instead of subtracting \((3z)^2\). Option D does not expand to the required four-term expression and omits the correct squared structure. The order of the two factors may be reversed without changing the product, but their signs must remain as shown. Memory cue: make the perfect square first, then use difference of squares.