Factroise: $4 a^{2} + 9 b^{2} + c^{2} - 12 a b - 6 b c + 4 a c$

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Solution

Formula: $(a + b - c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b - 2 b c - 2 c a$

Given,

$4 a^{2} + 9 b^{2} + c^{2} - 12 a b - 6 b c + 4 a c$

$= (2 a)^{2} + (3 b)^{2} + (1 c)^{2} - 2 (2 a) (3 b) - 2 (3 b) (1 c) + 2 (2 a) (1 c)$

$= (2 a + 3 b - c)^{2}$

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Similar questions

4 a^{2} + 9 b^{2} + c^{2} - 12 a b - 6 b c + 4 a c

4 a^{2} + 9 b^{2} + c^{2} - 12 a b + 6 b c - 4 a c

4 a^{2} + 9 b^{2} + 16 c^{2} + 12 a b - 24 b c - 16 a c

Factorise Completely : $4 a^{2} - 12 a b + 9 b^{2} + 4 a - 6 b$

4 a^{2} + 9 b^{2} - c^{2} + 12 a b = 0

a x + b y + c = 0

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