The value of $p$ which will make $2 m^{2} + p m n + 3 n^{2} - 5 n - 2$ equivalent to the product of two linear factors is

2

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\pm 3

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\pm 7

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- 4

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Solution

The correct option is C $\pm 7$
$3 n^{2} + n (p m - 5) + 2 m^{2} - 2 = 0$ (to find the factors) implies
$n = \frac{5 - p m \pm \sqrt{(p m - 5)^{2} - 12 (2 m^{2} - 2)}}{6}$
Now $D i s c r i m i n a n t = (p m - 5)^{2} - 12 (2 m^{2} - 2)$
$= p^{2} m^{2} - 10 p m + 25 - 24 m^{2} + 24$
$= m^{2} (p^{2} - 24) - 10 p m + 49$
$B^{2} - 4 A C = 0$ implies
$100 p^{2} - 4 (49) (p^{2} - 24) = 0$
$25 p^{2} - 49 (p^{2} - 24) = 0$
$p^{2} (25 - 49) + (49 \times 24) = 0$
$- 24 p^{2} + (49 \times 24) = 0$
$- p^{2} + 49 = 0$
Or
$p^{2} = 49$
Or
$p = \pm 7$

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