Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

tien.lam12

  • 4 years ago

What are the possible number of positive, negative, and complex zeros of f(x) = -3x4 - 5x3 - x2 - 8x + 4 ? A) Positive: 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0 B) Positive: 1; Negative: 3 or 1; Complex: 2 or 0 C) Positive: 3 or 1; Negative: 1; Complex: 2 or 0 D) Positive: 4 or 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0

  • This Question is Closed
  1. ChukRock
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    the numbers after the x's are exponents, correct?

  2. tien.lam12
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

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

    there are 2 positive and 2 complex roots

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

    i've tried using descartes rule of signs but i got confused to be honest so i solved it

  5. sahil51993
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    one root lie b/w 0 and 1

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

    right

  7. sahil51993
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    other will be negative as f(x) is increasing in R+

  8. sahil51993
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Am i right????

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

    no 2 positive

  10. sahil51993
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah 1 will be positive and one will not be possitive it can also be possilbe that roots may be irrational becuase irrational roots also exist in pairs

  11. estudier
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 5

    there is 1 sign change so by DRS there is 1 positive root Sub -x for x to get negative roots, 3 sign changes so 3 or 1 negative roots Complex roots are total roots, 4, less positive and negative roots so 0 or 2 of those. Answer is B

  12. 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