Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

itsjustme_lol

  • one year ago

Im having trouble with factoring..

  • This Question is Closed
  1. itsjustme_lol
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    im suppose to factor x^2+4 is this prime?

  2. itsjustme_lol
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    or (x+2)(x+2)

  3. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    it will have complex factors. else its prime.

  4. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    its NOT (x+2)(x+2)

  5. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[a^2+b^2 = (a+ib)(a-ib)\]

  6. itsjustme_lol
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok wait a second

  7. itsjustme_lol
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    (x + 2)(x – 2) (x + 2)(x + 2) (x + 1)(x + 4) ao it couldnt be any of these? im so confused with this one question

  8. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    no, it cannot be any of these. its prime, if we are considering only real factors.

  9. itsjustme_lol
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok. thankyou. i thought so, just was not sure. :) i appreicaite that.

  10. whpalmer4
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[(x+2)(x-2) = x^2 +2x - 2x -4 = x^2 -4\]\[(x+2)(x+2) = x^2 + 2x + 2x + 4 = x^2 + 4x + 4\]\[(x+1)(x+4) = x^2 + x + 4x + 4 = x^5 + 5x + 4\]

  11. whpalmer4
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    That i in there in the (a+ib)(a-ib) does the trick of reversing the sign of the final term: \[(a+ib)(a-ib) = a^2 + aib - aib -i^2b^2 = a^2 - i^2b^2\]but \[ i^2 = -1\] so \[a^2-i^2b^2=a^2-(-b^2)=a^2+b^2\]

  12. hartnn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    welcome ^_^

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

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.