Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

caroo.salinaas19

  • 3 years ago

Form a polynomial f(x) with real coefficients having the given degree and zeros. Degree: 4; zeros: 2i and -3i

  • This Question is Closed
  1. KingGeorge
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    In general, if you have a polynomial with real coefficients, and some complex root \(\alpha i\) where \(\alpha\) is a real number, the you also have the complex root \(-\alpha i\). Using this, can you tell me what all the roots of your polynomial will be?

  2. caroo.salinaas19
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    2i, -2i, -3i, and 3i?

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

    Bingo. So that means your factored polynomial will be \[(x-2i)(x+2i)(x-3i)(x+3i).\]Now you just have to expand it out.

  4. caroo.salinaas19
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i got this answer: x^3 -3ix^2-2ix^2+6ix^2-4xi^2+12i^2.

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

    Hmmm. I've definitely got something different. I'll walk you through the first few steps I did. \[(x-2i)(x+2i)=x^2-2ix+2ix-(2i)^2=x^2-4(-1)=x^2+4\]Note that \(i^2=-1\) by definition. Similarly, \[(x-3i)(x+3i)=x^2-3ix+3ix-(3i)^2=x^2-9(-1)=x^2+9\]Using this, can you find \[(x^2+4)(x^2+9)\]on your own?

  6. 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