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What is the complete factorised form of (4x^2+4xy+y^2-9z^2)?

Answer and explanation

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

The first three terms are a perfect square because \(4x^2=(2x)^2\), \(y^2=y^2\), and the middle term is \(4xy=2(2x)(y)\). Thus \(4x^2+4xy+y^2=(2x+y)^2\). The last term, \(9z^2\), is \((3z)^2\), so the expression is a difference of squares.

Using \(A^2-B^2=(A-B)(A+B)\), with \(A=2x+y\) and \(B=3z\), gives \((2x+y-3z)(2x+y+3z)\). This is the complete factorisation, so option A is correct. Option C stops before splitting the difference of squares and is only a partial form. Option B changes the sign of the \(4xy\) term, and option D uses the wrong square root of \(9z^2\).

Related tags

GroupingDifference Of SquaresFactorisation

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