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

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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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Other answers:

okay let me see...
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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