If $x^{2} + 2 x + 2 x y + m y - 3 = 0$ has two rational factors, then the values of m will be?

- 6, - 2

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- 6, 2

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6, - 2

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6, 2

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Solution

The correct option is C $6, - 2$
$x^{2} + 2 x + 2 x y + m y - 3 = 0$
$\Rightarrow x^{2} + 2 (1 + y) x + (m y - 3) = 0$
$\Rightarrow Δ = b^{2} - 4 a c > 0$
$\Rightarrow 4 {(1 + y)}^{2} - 4 \times 1 \times (m y - 3) > 0$
$\Rightarrow 1 + y^{2} + 2 y - m y + 3 > 0$
$\Rightarrow y^{2} + (2 - m) y + 4 > 0$
$\Rightarrow {(2 - m)}^{2} - 4 \times 1 \times 4 = 0$
$\Rightarrow {(2 - m)}^{2} - 16 = 0$
$\Rightarrow {(m - 2)}^{2} - 16 = 0$
$\Rightarrow (m - 2 - 4) (m - 2 + 4) = 0$
$\Rightarrow (m - 6) (m + 2) = 0$
$∴ m = 6, - 2$

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