Factorise:
$4 x^{2} + 8 x + 4$

4 (x - 1)^{2}

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4 (x + 1)^{2}

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2 (x + 1)^{2}

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2 (x - 1)^{2}

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Solution

The correct option is B $4 (x + 1)^{2}$
$4 x^{2} + 8 x + 4$
$= 4 (x^{2} + 2 x + 1)$
$= 4 (x^{2} + 2 \times x \times 1 + 1^{2})$
$= 4 (x + 1)^{2}$ [Since, $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ ]
$∴ 4 x^{2} + 8 x + 4 = 4 (x + 1)^{2}$

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