How do you factor $2 y^{4} + 3 y^{3} - 9 y^{2}$ $2y^4 + 3y^3 - 9y^2$ ?

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How do you factor $2 y^{4} + 3 y^{3} - 9 y^{2}$ $2y^4 + 3y^3 - 9y^2$ ?

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Answer 1 · 2015-05-08T23:06:21

You can factor $2 y^{4} + 3 y^{3} - 9 y^{2}$ $2y^4 + 3y^3 - 9y^2$ by grouping out $y^{2}$ $y^2$ .
The $y^{2}$ $y^2$ is present in all terms, so you can pull it out like so:

$y^{2} (2 y^{2} + 3 y - 9)$ $y^2(2y^2+3y-9)$

Answer 2 · 2015-05-08T23:07:15

$2 y^{4} + 3 y^{3} - 9 y^{2}$ $2y^4+3y^3-9y^2$

Extract the obvious common factor from each term
$y^{2} (2 y^{2} + 3 y - 9)$ $y^2(2y^2+3y-9)$

Then factor the quadratic
$y (2 y - 3) (y + 3)$ $y(2y-3)(y+3)$

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How do you factor $2 y^{4} + 3 y^{3} - 9 y^{2}$ $2y^4 + 3y^3 - 9y^2$ ?

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