The L.C.M. of $9 x^{2} + 6 x + 1, 3 x^{2} + 7 x + 2$ and $2 x^{2} + 3 x - 2$ is $(x + a) (2 x - b) (3 x + c)^{2}$ . Then $a - b - c =$ .

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Solution

The correct option is C 0.0
First expression:
$9 x^{2} + 6 x + 1 = (3 x + 1)^{2}$
Second expression:
$3 x^{2} + 7 x + 2 = 3 x^{2} + 6 x + x + 2 = 3 x (x + 2) + 1 (x + 2) = (3 x + 1) (x + 2)$
Third expression: $2 x^{2} + 3 x - 2 = 2 x^{2} + 4 x - x - 2 = 2 x (x + 2) - 1 (x + 2) = (2 x - 1) (x + 2)$
$∴$ Required L.C.M. is $(x + 2) (2 x - 1) (3 x + 1)^{2}$

$⟹ a = 2, b = 1, c = 1$

So, $a - b - c = 0$

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