How do you factor $3y^4-2y^2-5$ completely?

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Answer 1 · 2018-06-13T10:50:29

$3 y^4 - 2 y^2 -5 = (y^2+1)(3 y^2 - 5)$

$3 y^4 - 2 y^2 -5$

$= 3 y^4 + 3 y^2 - 5 y^2-5$

$= 3 y^2(y^2+1) - 5( y^2+1)$

$= (y^2+1)(3 y^2 - 5)$ [Ans]

How do you factor 3y^4-2y^2-53y^4-2y^2-5 completely?