If the expression $y^{2} + 2 x y + 2 x + m y - 3$ can be resolved into two rational factors, then $m$ must be

any possible real number

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any negative real number

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- 2

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3

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Solution

The correct option is C $- 2$
Treating it as a quadratic in $y$ , we get
$y^{2} + y (2 x + m) + (2 x - 3) = 0$
$D = (2 x + m)^{2} - 4 (2 x - 3)$
$= 4 x^{2} + 4 m x + m^{2} - 8 x + 12$
$= 4 x^{2} + 4 x (m - 2) + m^{2} + 12$
$= [4 x^{2} + 4 x (m - 2) + (m - 2)^{2}] + (m^{2} + 12) - (m - 2)^{2}$
$= [4 x^{2} + 4 x (m - 2) + (m - 2)^{2}] + [12 + 4 m - 4]$
$= (2 x + m - 2)^{2} + [12 + 4 m - 4]$
Hence,
$4 m = - 8$
$\Rightarrow m = - 2$ .

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