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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
okay let me see...
kayyy
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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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