Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

LovinBachata Group Title

Find The Four Real Zeros Of The Polynomial. f(x)= x^4+6x^3-15x^2-152x-240

  • one year ago
  • one year ago

  • This Question is Closed
  1. mathmate Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Descartes rule of signs tells us that there is only one positive root. Factoring theorem tells us that the rational roots would be factors 240 (which has many). To guess the positive root, seeing that the first two terms are positive, with small coefficients, and the remaining (negative) have large coefficients, we can try large factors, such as 4 (4^4=256), or 5 (5^4=625) by evaluating f(4) or f(5). Once we find the first one, the rest are negative roots of the remaining cubic.

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

    so the anser would be -4,-4,-3,5 ?

    • one year ago
  3. mathmate Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes, how did you find them?

    • one year ago
  4. LovinBachata Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i first used synthetic division using the number 4 and -4 and kept dividing until i had four solutions

    • one year ago
  5. mathmate Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Yep, that's great! Not everyone is familiar with synthetic division. How did you get to start with 4 and -4?

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

    Synthetic division is easier for me to use and because you gave me those numbers

    • one year ago
  7. mathmate Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I hope you'd be able to find those starter numbers in an exam using Descartes rule of signs and the factor theorem!

    • one year ago
  8. LovinBachata Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    I Hope So Tooo! Thank You Soo Much

    • one year ago
  9. mathmate Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    You synthetic division capability is a great asset in solving equations. Another way is to evaluate f(x) by nesting, for example, for f(5), you can almost do it by inspection (mentally): f(5)=(((5+6)*5-15)*5-152)*5-240 =((55-15)*5-152)*5-240 =(200-152)*5-240 =48*5-240 =0

    • one year ago
  10. LovinBachata Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    but how do you know with what numbers to start with?

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

    Assuming you have a starter number of 5, then evaluating using nesting (as shown above) is fast, accurate and less error prone... with a little practice. The starter number remain guesses, with he help of Descartes Theorem and Factor theorem. Are you already familiar with these two?

    • one year ago
  12. LovinBachata Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    im only familiar with the factor theorem but i dont remember Descartes theroem do you usuallly look ath the last numbeer?

    • one year ago
  13. mathmate Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    If the leading coefficient is 1, then only the last number counts. If the leading coefficient id different from one, then the possible factors are of the form (ax+b) where a is a factor of the leading coefficient, and b is a factor of the last number (constant term). Descartes rule tells us that the maximum number of positive roots is equal to the number of sign changes. In the above case, sign only changed once from + to -, so there is only one positive root, and a maximum of three negative roots. Maximum means 2,4,6... can be replaced by complex roots, since complex roots always come in pairs. For example, in our case, there is one positive root (cannot be replaced by two complex), and one or three real negative roots. Since the question says 4 real roots, we can count on 3 negative (real) roots.

    • one year ago
  14. mathmate Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    * replace (ax+b) by \( (ax\pm b)\)

    • one year ago
  15. LovinBachata Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    so the fact that there is one postive sign means that the solutions would have one positive solution?

    • one year ago
  16. mathmate Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    It's the \( change \) in sign that counts. x^2-2x+1=0 has two positive roots, but x^2-x-2=0 has only one.

    • one year ago
    • Attachments:

See more questions >>>

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.