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

Answer and explanation

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

The first three terms are the perfect square \(x^2+2xy+y^2=(x+y)^2\). The expression becomes \((x+y)^2-z^2\), which is a difference of squares. Using \(u^2-v^2=(u-v)(u+v)\), with \(u=x+y\) and \(v=z\), gives \((x+y-z)(x+y+z)\). Therefore option A is correct.

Expanding confirms it: \((x+y-z)(x+y+z)=(x+y)^2-z^2=x^2+2xy+y^2-z^2\). Option C is an equivalent intermediate form but not the complete factorisation. Option B changes \(x+y\) to \(x-y\), altering the middle term, and D does not reproduce the expression. Recognising a perfect square before applying difference of squares makes the process short and reliable.

Related tags

Combined FactorisationPerfect SquareDifference Of Squares

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