Using apporpriate identity, factorize $9 x^{2} + 6 x + 1$ .

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Solution

$9 x^{2} + 6 x + 1$
$= 3^{2} x^{2} + 6 x + 1^{2}$
$= {(3 x)}^{2} + 6 x + 1^{2}$
$= {(3 x)}^{2} + 2 (3 x) (1) + 1^{2}$
Using identity: ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$
whereas, $a = 3 x, b = 1$
$= {(3 x + 1)}^{2}$
$= (3 x + 1) (3 x + 1)$
Hence $9 x^{2} + 6 x + 1 = (3 x + 1) (3 x + 1)$

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