How do you factor $6 t^{4} + t^{2} - 12$ $6t^4 + t^2 - 12$ ?

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How do you factor $6 t^{4} + t^{2} - 12$ $6t^4 + t^2 - 12$ ?

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Answer 1 · 2016-04-22T10:51:07

$(\sqrt{3} \cdot t + 2) (\sqrt{3} \cdot t - 2) (2 t^{2} + 3)$ $(sqrt3*t+2)(sqrt3*t-2)(2t^2+3)$

Explanation:

$6 t^{4} + t^{2} - 12 = 6 t^{4} + 9 t^{2} - 8 t^{2} - 12 = 3 t^{2} (2 t^{2} + 3) - 4 (2 t^{2} + 3) = (3 t^{2} - 4) (2 t^{2} + 3) = (\sqrt{3} \cdot t + 2) (\sqrt{3} \cdot t - 2) (2 t^{2} + 3)$ $6t^4+t^2-12 =6t^4+9t^2-8t^2-12=3t^2(2t^2+3)-4(2t^2+3) =(3t^2-4)(2t^2+3)=(sqrt3*t+2)(sqrt3*t-2)(2t^2+3)$ [Ans]

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How do you factor $6 t^{4} + t^{2} - 12$ $6t^4 + t^2 - 12$ ?

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