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

Question

2483 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-13T10:50:29

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

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

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

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

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

How do you factor 3y^4-2y^2-53y4−2y2−53y^4-2y^2-5 completely?