A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

Solve the equation 2(x^4)+3(x^3)-7(x^2)+3x-9=0 by first finding any rational roots. Thanks!

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

    possible rational roots are \[\pm1, \pm3, \pm9, \pm\frac{1}{2}, \pm\frac{3}{2},\pm\frac{9}{2}\]

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

    a lot to check actually so i would start with the easy ones : 1 and -1

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

    1 does not work

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

    if you give me a minute i will find them for you

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

    sure.

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

    -1 does not work either

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

    don't you need to use synthetic division to find the other roots as well?

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

    well you need to find at least one to start. you can use synthetic division sure. i was going to cheat

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

    -3 works

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

    once you have -3 you can use synthetic division to factor

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

    and then what do I do w/ the depressed equation?

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

    depressed?

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

    you will get \[p(x)=(x+3)q(x)\]

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

    you find \[q(x)\] by dividing

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

    in fact i have done it.

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

    would you like the answer or do you want to find it?

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

    both. if you do not mind.

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

    ok. first of all \[p(-3)=0\] you can check this by synthetic division, or by replacing x by -3 which is kind of a pain

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

    ok, i have done the division. the equation is 2x^3-3x^2+2x-3=0.

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

    k you were ahead of me

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

    now the possible rational roots have narrowed somewhat. we know 1 and -1 don't work. we found -3. 3 does not work. now you can try \[\frac{3}{2}\] and \[-\frac{3}{2}\]

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

    i give you a hint, it is not the negative one!

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

    try 3/2 and you will get no remainder

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

    yes, i found that.

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

    when do I stop finding roots?

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

    ok and you get \[p(x)=(x+3)(x-\frac{3}{2})(2x^2+2)\]

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

    or \[p(x)=(x+3)(2x-3)(x^2+1)\]

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

    now you can stop because \[x^2+1\] does not factor over the real numbers

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

    if you are using complex numbers then i guess you can write \[p(x)=(x+3)(2x-3)(x+i)(x-i)\] but that is kind of silly because we are working with a real valued polynomial

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

    is it also b/c the original equation is to the third degree? somehow, I also found +i, -i...

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

    yes because the roots of \[x^2+1\] are i and -i

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

    we have found the 4 zeros, 2 real and two complex. of course you can stop now because a polynomial of degree 4 can have a most 4 zeros

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

    ohhhh, so +i and -i can b considered any number?

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

    oh no

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

    they have a specific meaning

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

    square root of -1.

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

    \[x^2+1=0\] \[x^2=-1\] \[x=\pm\sqrt{-1}\]

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

    it is just a symbol for \[\sqrt{-1}\]

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

    ahhhhhh. ok. thanks a ton!!

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

    not a real number

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

    welcome. hope it was clear

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

    definitely.

  43. 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

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.