Find the value of $m^{2}$ for which the expression $2 x^{2} + m x y + 3 y^{2} - 5 y - 2$ can be resolved into two rational linear factors

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Solution

We know that $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c$ can be resolved into two linear factors if and only if
$a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$
Given expression is
$2 x^{2} + m x y + 3 y^{2} - 5 y - 2$ ....(1)
Here, $a = 2, h = m / 2, b = 3, g = 0, f = - 5 / 2, c = - 2$ . Therefore, expression
$2 x^{2} + m x y + 3 y^{2} - 5 y - 2$ will have two linear factors if and only if
$a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$
or $2 \times 3 (- 2) + 2 (\frac{- 5}{2}) (0) (\frac{m}{2}) - 2 {(\frac{- 5}{2})}^{2} - 3 \times 0^{2} - (- 2) {(\frac{m}{2})}^{2} = 0$
or $- 12 - \frac{25}{2} + \frac{m^{2}}{2} = 0$ or $m^{2} = 49$ or $m = \pm 7$

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