The linear factor(s) of the equation $x^{2} + 4 x y + 4 y^{2} + 3 x + 6 y - 4 = 0$ is/are

x + 2 y + 4 = 0

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x + 2 y - 4 = 0

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x + 2 y - 1 = 0

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x + 2 y + 1 = 0

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Solution

The correct option is C $x + 2 y - 1 = 0$
$x^{2} + 4 x y + 4 y^{2} + 3 x + 6 y - 4 = 0$
Writing equation in terms of $y,$ we get
$4 y^{2} + (4 x + 6) y + (x^{2} + 3 x - 4) = 0$
Now, using quadratic formula, we have
$y = \frac{- (4 x + 6) \pm \sqrt{(4 x + 6)^{2} - 4 \cdot 4 \cdot (x^{2} + 3 x - 4)}}{2 \cdot 4}$
$\Rightarrow y = \frac{- (4 x + 6) \pm 10}{2 \cdot 4}$
$\Rightarrow y = \frac{- (2 x + 3) \pm 5}{4} \Rightarrow y = \frac{- x + 1}{2} or y = \frac{- x - 4}{2} \Rightarrow x + 2 y - 1 = 0 or x + 2 y + 4 = 0$

Alternate Solution:
$x^{2} + 4 x y + 4 y^{2} + 3 x + 6 y - 4 = 0 \Rightarrow (x^{2} + 4 x y + 4 y^{2}) + (3 x + 6 y) - 4 = 0 \Rightarrow (x + 2 y)^{2} + 3 (x + 2 y) - 4 = 0 \Rightarrow ((x + 2 y) + 4) ((x + 2 y) - 1) = 0 \Rightarrow x + 2 y + 4 = 0 or x + 2 y - 1 = 0$

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