how do i factor this out completely? 12y^5-34xy^4+14x^2y^3

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how do i factor this out completely? 12y^5-34xy^4+14x^2y^3

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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You can first factor out 2y^3 and you get: 2y^3(6y^2-17xy+7x^2) Rearranging this slightly only because I like putting x first and because the middle term -17xy has x first: 2y^3(7x^2-17xy+6y^2)-- Now (7x^2-17xy+6y^2) can also be factored: 2y^3(7x-3y)(x-2y) This is the factor
\[2y ^{3}[(2y-x)(3y-7x)]\]

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