One of the factors of $12 a^{2} - 31 a b + 20 b^{2}$ is $(3 a - 4 b)$ .
Therefore, the other factor is

(4 a - 5 b)

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(3 a + 4 b)

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(4 a - 3 b)

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(2 a - 3 b)

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Solution

The correct option is A $(4 a - 5 b)$
$12 a^{2} - 31 a b + 20 b^{2}$
We need to find two integers whose product is $240$ and sum is $- 31$ .
The factors are $- 16$ and $- 15$ .

$∴ 12 a^{2} - 31 a b + 20 b^{2}$
$= 12 a^{2} - 16 a b - 15 a b + 20 b^{2}$
$= 4 a (3 a - 4 b) - 5 b (3 a - 4 b)$
$= (3 a - 4 b) (4 a - 5 b)$

We already know that $(3 a - 4 b)$ is a factor.
So, the other factor is $(4 a - 5 b)$ .

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