Factorize $4 x^{2} + y^{2} + 9 z^{2} - 4 x y + 6 y z - 12 z x$

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Solution

Given
$4 x^{2} + y^{2} + 9 z^{2} - 4 x y + 6 y z - 12 z x$
Since, $a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a = (a + b + c)^{2}$
Substitute $a = 2 x, b = - y, c = - 3 z$ , we get
$(2 x)^{2} + (- y)^{2} + (- 3 z)^{2} + 2 (2 x) (- y) + 2 (- y) (- 3 z) + 2 (- 3 z) (2 x) = [(2 x) + (- y) + (- 3 z)]^{2}$
$= (2 x - y - 3 z)^{2} = (2 x - y - 3 z) (2 x - y - 3 z)$
Hence, the factors of the given expression are $(2 x - y - 3 z) (2 x - y - 3 z)$

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