Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

kitsune0724

  • 3 years ago

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

  • This Question is Closed
  1. ramya25
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    substitute t = n^2

  2. kitsune0724
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    My teacher wants us to use either factoring or the quadratic formula on these problems.

  3. ramya25
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes. first you have to change it to quadratic then you can factor

  4. kitsune0724
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay let me see...

  5. ramya25
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    kayyy

  6. mukushla
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    use this hint \[\large 3n^4-4n^2+1=3n^4-3n^2-(n^2-1)=3n^2(n-1)-(n-1)(n+1)\]

  7. ramya25
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes grouping gives quick factors here. check the steps of @mukushla

  8. kitsune0724
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thank y'all.

  9. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy