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evy15

Find all the zeros of the eq. 2x^4-5x^3+53x^2-125x+75=0

  • one year ago
  • one year ago

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  1. mathmate
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    This one can be factorized, so you will need to use the following tools: Descartes rule of signs, factorization theorem, fundamental theorem of algebra Are any or some of these famililar to you? Also, this may indicate that you or your teacher have a typo in the other problem. Please re-check for typos on the other problem.

    • one year ago
  2. mathmate
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    To make things simpler, when the sum of coefficients = 0, it means that (x-1) is a factor.

    • one year ago
  3. evy15
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    no I dnt know those methods

    • one year ago
  4. mathmate
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    Descartes rule of signs tells us that we have either 4 real positive roots, or two real positive roots and two complex roots, or 4 complex roots.

    • one year ago
  5. evy15
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    ok

    • one year ago
  6. mathmate
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    http://en.wikipedia.org/wiki/Descartes'_rule_of_signs http://en.wikipedia.org/wiki/Factor_theorem http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra Since you have been missing school for at least one day, it may well be advisable to catch up with some of these topics. Hope you got notes from your friends though.

    • one year ago
  7. mathmate
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    So is (x-1) a factor to the equation?

    • one year ago
  8. evy15
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    yes

    • one year ago
  9. mathmate
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    So you can do a synthetic division, or a long division to reduce the quartic to a cubic, i.e. find (2*x^4-5*x^3+53*x^2+(-125)*x+75)/(x-1)=?

    • one year ago
  10. evy15
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    2x^3-3x^2+50x-75

    • one year ago
  11. mathmate
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    Good, now we can rule out the third case (4 complex roots).

    • one year ago
  12. evy15
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    ok

    • one year ago
  13. mathmate
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    Using the factor theorem, you will need to try factors of the form (ax+b) where a=1 or 2 (from the leading coefficient 2), and b=+/- all possible factors of 75, which are 1,3,5,15,25,75, i.e. 2*2*6=24 possibilities. Once you find that, you can again do the synthetic division or long division to reduce the cubic to a quadratic.

    • one year ago
  14. evy15
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    what do I divide it by

    • one year ago
  15. mathmate
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    When you find another root (i.e. another factor) of the cubic.

    • one year ago
  16. evy15
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    oh ok, can I use the calculator for that

    • one year ago
  17. KingGeorge
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    Sorry to bust in here, but \(2x^3-3x^2+50x-75\) is relatively easily factored by grouping, and thus we can skip the whole guess and check portion of finding roots.

    • one year ago
  18. mathmate
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    In fact, Descartes rule of sign says we don't have to try factors of the form (a+b), only those of the form (a-b) because all real roots are positive.

    • one year ago
  19. mathmate
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    So that reduces to 12 possibilities only.

    • one year ago
  20. mathmate
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    Very true, thanks KingGeorge!

    • one year ago
  21. KingGeorge
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    You group like so \[2x^3-3x^2+50x-75=x^2(2x-3)+25(2x-3)\]Can evy15 finish the factoring from here?

    • one year ago
  22. mathmate
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    @evy15 are you able to complete the solution from here?

    • one year ago
  23. evy15
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    can you guide me to what to do next

    • one year ago
  24. evy15
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    ok hold on

    • one year ago
  25. evy15
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    x=i+-5 and 3/2

    • one year ago
  26. evy15
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    i mean =-5i

    • one year ago
  27. mathmate
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    Good job, (+\- 5i). Now you have all 4 zeroes!

    • one year ago
  28. mathmate
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    Can you enumerate the four zeroes?

    • one year ago
  29. evy15
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    so its 1,+/-5i, and 3/2?

    • one year ago
  30. evy15
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    or-1

    • one year ago
  31. mathmate
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    Excellent! There you are! Not -1. Your list is good.

    • one year ago
  32. mathmate
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    Big thanks to KingGeorge who cut the work in half!

    • one year ago
  33. evy15
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    yes thank you guys so much

    • one year ago
  34. mathmate
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    yw! :)

    • one year ago
  35. KingGeorge
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    As an addendum, it's possible to factor the original function by grouping if you're clever. \[\begin{aligned} 2x^4-5x^3+53x^2-125x+75&=2x^4-5x^3+3x^2+50x^2-125x+75 \\ &=x^2(2x^2-5x+3)+25(2x^2-5x+3) \\ &=(x^2+25)(2x^2-5x+3)\\ &=(x^2+25)(x-1)(2x-3) \end{aligned}\]

    • one year ago
  36. KingGeorge
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    But this is not an obvious, nor the easiest, way to do it.

    • one year ago
  37. mathmate
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    Yep, I will look more carefully for these cases in the future.

    • one year ago
  38. evy15
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    can i do the same with x^5-3x^4-24x^3-72x^2+12x=12

    • one year ago
  39. evy15
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    sorry wrote the wrong one

    • one year ago
  40. evy15
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    x^5-3x^4-24x^3-72x^2-25x+75=0

    • one year ago
  41. KingGeorge
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    That can definitely be factored by grouping. However, instead of having two groups like I did before, we'll have three groups.

    • one year ago
  42. KingGeorge
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    Your first group will be \[x^4(x-3).\]Knowing this, can you give me the rest of the factorization?

    • one year ago
  43. evy15
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    x^4(x-3)-24x^2(x-3)-25(x-3)

    • one year ago
  44. KingGeorge
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    Perfect. And can you finish the rest of the factorization?

    • one year ago
  45. evy15
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    would it be (x^4-24x^2-25) and (x-3)?

    • one year ago
  46. KingGeorge
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    Right again. Now you just need to factor \(x^4-24x^2-25\). You can do this by pretending it's a quadratic equation, and not a quartic. So how does \(x^2-24x-25\) factor?

    • one year ago
  47. evy15
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    x-25 and x+1

    • one year ago
  48. KingGeorge
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    Perfect. So that means \(x^4-24x^2-25=(x^2-25)(x^2+1)\). Hence, \[x^5-3x^4-24x^3-72x^2-25x+75=(x^2-25)(x^2+1)(x-3)\]From here, you just have some sums/differences of squares, and should be able to figure out the roots. I've got to go now, but you seem to be catching the hang of this!

    • one year ago
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