Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

yociyoci

  • 2 years ago

Solve x^4+x^3+x^2+x+1 = 0 by hand

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

    You're studying complex numbers I assume. I will give you a hint: \[x^4+x^3+x^2+x+1=\frac{x^5-1}{x-1}.\] Solve it now by first finding the roots of unity of \(x^5=1\).

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

    Why is x^4+x^3+x^2+x+1 = (x^5-1)/(x-1)

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

    This come from the factorization \(x^5-1=(x-1)(x^4+x^3+x^2+x+1)\). If you don't already know that. Then you can easily notice that \(1\) is a root of \(x^5-1\), and then you can use synthetic division to get the above factorization.

  4. Mr.Math
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    comes*

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

    o right

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

    but does that give you all of the roots?

  7. Mr.Math
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    \(x^5-1\) has all the roots of your problem and one more root which is \(x=1\), so you have to exclude it.

  8. Mr.Math
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    You know how to solve \(x^5=1\)?

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

    1 and 4 complex solutions

  10. Mr.Math
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Exactly. The four complex solutions are what you're looking for, as you can see from the factorization of \(x^5-1\).

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

    im not really sure about how to solve x^5=1

  12. Mr.Math
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Have a look at this http://www.youtube.com/watch?v=9Z1uNz__2xA

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

    o ty

  14. moneybird
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Or First divide the entire equation by x^2 \[x^2+x+1+\frac{1}{x}+\frac{1}{x^2}=0 \implies x^2+\frac{1}{x^2}+x+\frac{1}{x}+1=0\] Let u = x+1/x \[u^2 = x^2+2+\frac{1}{x^2}\] The original equation now becomes: \[u^2-2+u+1 = 0 \implies u^2+u-1=0\] Use quadratic equation then substitute u = x+1/x back and then solve the quadratic again.

  15. 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.