Question

Factorise this doddle question

Answer 1

$4 {x}^{2} + 9 {y}^{2} + 12 {z}^{2} + 12xy - 24yz - 16xz$

$(2 {x})^{2} + (3 {y})^{2} + ( - 4 {z})^{2} + (2 \times 2x \times 3y) + (2 \times 3y \times - 4z) + (2 \times - 4z \times 2x)$

$= >( 2x + 3y - 4z{})^{2}$

$= > (2x + 3y - 4z) \: (2x + 3y - 4z)$

$using \: identity \: where \: {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ac = ( a + b + c{})^{2}$

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