The complete factor of $12 a^{4} - 3 a^{2} - 54$ is

3 (2 a + 3) (2 a - 3) (a^{2} + 2)

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(2 a + 3) (2 a - 3) (a^{2} + 2)

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3 (2 a + 3) (2 a - 3) (a + 2)

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3 (2 a + 3)^{2} (a^{2} + 2)

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Solution

The correct option is A $3 (2 a + 3) (2 a - 3) (a^{2} + 2)$
$12 a^{4} - 3 a^{2} - 54$
Take $3$ as common, we get
$= 3 (4 a^{3} - a^{3} - 18)$
$= 3 (4 a^{2} - 9) (a^{2} + 2)$
$= (4 a^{2} - 9)$ as a binomial that can be factored into $(2 a + 3) (2 a - 3)$ .
So the completely factored result is $3 (2 a + 3) (2 a - 3) (a^{2} + 2)$

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