Factorise : $4 x^{2} + 4 x + 1$

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Solution

Given: $4 x^{2} + 4 x + 1$
$= 4 x^{2} + 2 x + 2 x + 1$ , (split middle term so that resultant is same)
$= (4 x^{2} + 2 x) + (2 x + 1)$ , (group pair of terms)
$= 2 x (2 x + 1) + 1 (2 x + 1)$ , (factor each binomials)
$= (2 x + 1) (2 x + 1)$ , (factor out common factor $(2 x + 1)$ )
$= (2 x + 1)^{2}$

Aliter, use $(a + b^{2}) = a^{2} + 2 a b + b^{2}$
$\Rightarrow 4 x^{2} + 4 x + 1 = (2 x)^{2} + 2 \times 2 x + 1^{2} = (2 x + 1)^{2}$

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