How would solve 3n^4-4n^2+1=0 by factoring?

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How would solve 3n^4-4n^2+1=0 by factoring?

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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substitute t = n^2
My teacher wants us to use either factoring or the quadratic formula on these problems.
Yes. first you have to change it to quadratic then you can factor

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okay let me see...
kayyy
use this hint \[\large 3n^4-4n^2+1=3n^4-3n^2-(n^2-1)=3n^2(n-1)-(n-1)(n+1)\]
Yes grouping gives quick factors here. check the steps of @mukushla
thank y'all.

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