Question : The factors of $x^2+4 y^2+4 y-4 x y-2 x-8$ are:
Option 1: $(x-2 y-4)(x-2 y+2)$
Option 2: $\left(x^2-2 y-4\right)\left(x^2-2 y+2\right)$
Option 3: $(x+2 y-4)(x+2 y+2)$
Option 4: $\left(x^2-2 y-4\right)\left(x^2+2 y+2\right)$
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Correct Answer: $(x-2 y-4)(x-2 y+2)$
Solution :
$x^2+4 y^2+4 y-4 x y-2 x-8$
= $(x^2+(2y)^2-2(2xy))-2(x-2y)-8$
= $(x-2y)^2-2(x-2y)-8)$
= $(x-2y)^2-4(x-2y)+2(x-2y)-8)$
= $(x-2y+2)(x-2y-4)$
Hence, the correct answer is $(x-2y+2)(x-2y-4)$.
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