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Which is the correct factorisation of (x^2+2xy+y^2-9z^2)?

Answer and explanation

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

The first three terms form a perfect square because they match the identity \(u^2+2uv+v^2=(u+v)^2\). Taking \(u=x\) and \(v=y\), we get \(x^2+2xy+y^2=(x+y)^2\). The remaining term is \(9z^2=(3z)^2\), so the whole expression becomes the difference of the squares \((x+y)^2-(3z)^2\). The correct method is therefore to identify the perfect square first and then use the difference-of-squares identity.

Using \(A^2-B^2=(A-B)(A+B)\), we obtain \((x+y-3z)(x+y+3z)\), which is option B. Expanding confirms it: the product gives \((x+y)^2-9z^2\), and then \((x+y)^2=x^2+2xy+y^2\). Option A uses \(x-y\), which would change the sign of the middle term. Options C and D do not produce the required terms, so they cannot be correct.

Related tags

Mixed IdentityPerfect SquareDifference Of Squares

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