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evy15

  • one year ago

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

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  1. mathmate
    • one year ago
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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.

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

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

  4. mathmate
    • one year ago
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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.

  5. evy15
    • one year ago
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    ok

  6. mathmate
    • one year ago
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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.

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

  8. evy15
    • one year ago
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    yes

  9. mathmate
    • one year ago
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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)=?

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

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

  12. evy15
    • one year ago
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    ok

  13. mathmate
    • one year ago
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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.

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

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

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

  17. KingGeorge
    • one year ago
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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.

  18. mathmate
    • one year ago
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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.

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

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

  21. KingGeorge
    • one year ago
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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?

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

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

  24. evy15
    • one year ago
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    ok hold on

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

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

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

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

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

  30. evy15
    • one year ago
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    or-1

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

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

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

  34. mathmate
    • one year ago
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    yw! :)

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

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

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

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

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

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

  41. KingGeorge
    • one year ago
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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.

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

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

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

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

  46. KingGeorge
    • one year ago
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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?

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

  48. KingGeorge
    • one year ago
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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!

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