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If $x^{2} + 4 x y + 4 y^{2} + 4 x + c y + 3$ can be written as the product of two linear factors, then the value of ‘c’ is

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Detailed Solution

a = 1, h = 2, b = 4, g = 2, f = \frac{c}{2}, c = 3

$a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$

$a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$

$1.4.3 + 2 \frac{c}{2} .2.2 - 1. \frac{c^{2}}{4} - 4.4 - 3.4$

$= 0$

$12 + 4 c - \frac{c^{2}}{4} - 16 - 12 = 0$

$4 c - \frac{c^{2}}{4} - 16 = 0$

$c = 8$

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