Factorise: $4 a^{2} + 9 b^{2} + 16 c^{2} + 12 a b - 24 b c - 16 a c$ .

(2 a + 3 b + 4 c)^{2}

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(2 a - 3 b - 4 c)^{2}

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(2 a + 3 b - 4 c)^{2}

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(2 a - 3 b + 4 c)^{2}

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Solution

The correct option is C $(2 a + 3 b - 4 c)^{2}$
Given:

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

Use the formula:

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

Simplify the equation.

$4 a^{2} + 9 b^{2} + 16 c^{2} + 12 a b - 24 b c - 16 a c = {(2 a)}^{2} + {(3 b)}^{2} + {(- 4 c)}^{2} + 2 \times 2 \times 3 a b + 2 \times (- 4) \times 3 b c + 2 \times (- 4) \times 2 a c$

$= {(2 a + 3 b - 4 c)}^{2}$

Thus, the factor of the equation is ${(2 a + 3 b - 4 c)}^{2}$ .

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