Factorize the following
$14 x^{2} + 31 x + 6$

(14 x + 3) (x + 2)

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(14 x - 3) (x + 2)

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(14 x + 3) (x - 2)

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None of these

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Solution

The correct option is C $(14 x + 3) (x + 2)$
Given,
$14 x^{2} + 31 x + 6$

$28 + 3 = 31$ ...............the split for the middle term

$28 \times 3 = 84$ ...........the product for the first and the last term

Now splitting the middle term, we get

$= 14 x^{2} + 28 x + 3 x + 6$
$= 14 x (x + 2) + 3 (x + 2)$

$= (x + 2) (14 x + 3)$

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