Factorise: $4 x^{2} + 4 x y + y^{2}$

$(2 x - y)^{2}$

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$(4 x + y)^{2}$

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$(x + 2 y)^{2}$

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$(2 x + y)^{2}$

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Solution

The correct option is D
$(2 x + y)^{2}$

The expression $4 x^{2} + 4 x y + y^{2}$ can be written as,
$= (2 x)^{2} + 4 x y + y^{2}$ ...(i)
Using the identity, $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ , we get

$a^{2} = (2 x)^{2} \Rightarrow a = 2 x$

$b^{2} = y^{2} \Rightarrow b = y$

$\Rightarrow 4 x^{2} + 4 x y + y^{2} = (2 x + y)^{2}$

Therefore, the factors of $4 x^{2} + 4 x y + y^{2}$ are $(2 x + y)$ and $(2 x + y)$ .

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