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What is the factorised form of \(z^2-4z+4\)?

Answer and explanation

Correct answer: \((z-2)^2\)

The direct answer is C: \((z-2)^2\). Recognise the expression as a perfect-square trinomial. The first term is \(z^2\), the last term is \(4=2^2\), and the middle term is \(-4z=-2
times z imes2\). Therefore it matches \((u-v)^2=u^2-2uv+v^2\), with u=z and v=2: \(z^2-4z+4=(z-2)^2\). Option A gives \(z^2+4z+4\), so its middle sign is wrong. Option B is only a partially rearranged expression; expanding it gives \(z^2-4z+4\), but it is not the requested compact factorised product of the displayed options. Option D expands to \(z^2-3z-4\), which is different. Option C can be checked directly: \((z-2)(z-2)=z^2-2z-2z+4=z^2-4z+4\). Memory cue: when the first and last terms are squares, check whether the middle term is plus or minus twice their square roots multiplied together.

Related tags

FactorisationPerfect-Square TrinomialAlgebraic IdentitiesClass 9 MathematicsExploring Algebraic IdentitiesMathematicsClass 9 Mcq

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