Factorise : $x^{2} + 14 x + 49$

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Solution

Given that: $x^{2} + 14 x + 49$
$= x^{2} + 7 x + 7 x + 49$ , (split middle term so that resultant is same)
$= (x^{2} + 7 x) + (7 x + 49)$ , (group pair of terms)
$= x (x + 7) + 7 (x + 7)$ , (factor each binomials)
$= (x + 7) (x + 7)$ , (factor out common factor $(x + 7)$ )
$= (x + 7)^{2}$

Aliter, use $(a + b^{2}) = a^{2} + 2 a b + b^{2}$
$\Rightarrow x^{2} + 14 x + 49 = x^{2} + 2 \times 7 x + 7^{2} = (x + 7)^{2}$

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