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Can't find all the zeroes of a polynomial

  1. Oct 1, 2016 #1
    1. The problem statement, all variables and given/known data
    Help i have a homework quiz done and i simply can't find out how to do the 3rd problem as we haven't even learned how to do it or maybe my notes aren't good or something , however im close to an A in the class and this would help bring it closer. It asks me: "Find all the zeros of: X^4+3x-18". I have looked around online and only could find the rational theory to solve it( we haven't learned this) and it didnt even work. Im wondering if you guys could help me out.

    2. Relevant equations

    X^4+3x-18
    3. The attempt at a solution
    I tried factoring and stuff i learned in algerba 2 but thats all for quadratic equations how did Isaac Newton even figure this out??? I used an online calculator and was given the roots -2.229 and 1.875 but i dont know how to get those roots.
     
  2. jcsd
  3. Oct 1, 2016 #2

    fresh_42

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    There are closed formulas to solve this equation (first found by Lodovico Ferrari (1522–1565), published by Gerolamo Cardano 1545), but they are rather unpleasant. I assume the online solutions use numerical procedures, e.g. the Newton-Raphson method (around 1700) which uses an iteration with tangents.

    In some cases - not this one - one can guess a root and use polynomial division to reduce the degree.
     
  4. Oct 1, 2016 #3
    thank you. This will help me find the answer more. I can't believe my teacher gave me a problem like this.
     
  5. Oct 1, 2016 #4

    SammyS

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    A graphing calculator can give you those results.

    Are you sure that you copied the problem correctly?

    The polynomial, x4 + 3x2 - 18, can be factored.
     
  6. Oct 1, 2016 #5

    fresh_42

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    If you're interested in only the formulas, then I have a Wiki link for you. Unfortunately not in English, but the formulas don't bother and I'm too lazy to translate the entire entry.
    ... which is an indication that @SammyS is right and you made a copy and paste error with the formula.
     
  7. Oct 1, 2016 #6
    I dont have a phone on me, but thats the exact equation. Its literally unfactorable at my current level of math. Its x raised to the 4th plus 3x minus 18( x^4+3x-18)
     
  8. Oct 1, 2016 #7
    I just need to be able to find out how to do this problem. Im trying to get a hundred percent on this.
     
  9. Oct 2, 2016 #8

    fresh_42

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    Then have a look:
    It's so overabundant with formulas that the wrong language is only a minor difficulty.

    If you plot a graph you could see, how to apply Newton-Raphson:
    Start with any tangent near a root. Intersect it with the x-axis, compute the function value there, and compute a tangent there. Then intersect again with the x-axis and so on.
    (Edit: If I remember correctly. But the Wiki link to the method I posted above should be more precise.)
     
  10. Oct 2, 2016 #9

    ehild

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    You can try an iterative procedure: start with some x value (say, x = -1 or x = 1 ) and calculate ##x'=\sqrt{\sqrt{3(6-x)}}##. Then repeat with x=x' till x and x' are close enough together.
     
  11. Oct 2, 2016 #10

    Ray Vickson

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    Well, exact solutions are nasty in this case. For example, one of the two real roots is
    -1/12/(3+3^(1/2)*2051^(1/2))^(1/6)*(2^(7/12)*(-2^(1/2)*(2^(1/3)*3^(2/3)*(3+3^(1/2)*2051^(1/2))^(2/3)-48)^(1/2)*3^(2/3)*(3+3^(1/2)*2051^(1/2))^(2/3)+48*2^(1/6)*(2^(1/3)*3^(2/3)*(3+3^(1/2)*2051^(1/2))^(2/3)-48)^(1/2)+12*2^(1/6)*3^(1/2)*(3+3^(1/2)*2051^(1/2))^(1/2))^(1/2)+2^(2/3)*(2^(1/3)*3^(2/3)*(3+3^(1/2)*2051^(1/2))^(2/3)-48)^(3/4))*3^(5/6)/(2^(1/3)*3^(2/3)*(3+3^(1/2)*2051^(1/2))^(2/3)-48)^(1/4)

    Such formulas are hardly ever used in getting numerical answers; modern numerical root-finding algorithms are usually very fast and efficient and often more accurate (i.e, less susceptible to roundoff errors in computation).
     
  12. Oct 2, 2016 #11

    OmCheeto

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    It should be noted that "-2^(1/2)" is misinterpreted by some spreadsheets as: √(-2)
    To correct this, I used the form: -(2^(1/2))

    ps. I spent all day on this problem yesterday........ When I finally saw the solutions for the quartic equation, I was very happy to hear that even mathematicians won't go there.
     
    Last edited: Oct 2, 2016
  13. Oct 2, 2016 #12

    epenguin

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    Well just answer that you cannot see and there does not seem to be any simple factorisation involving integers for the polynomial you quoted, but on the other hand if the question is ... (the equation suggested in #4) then you can factorise in the following way...

    You might even get extra credit for that which after all would be earned because of the time you have spent on it. :oldsmile:
     
  14. Oct 2, 2016 #13
    thank you and all you guys i should do pretty good on this quiz.
     
  15. Oct 3, 2016 #14

    SammyS

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    Here's an alternative look.

    A degree 4 polynomial with real coefficients has either 0 or 2 or 4 real zeros. (We know from the graph that the polynomial discussed here has 2 real zeros.) We can factor this polynomial into to quadratics, each with real coefficients. Thus we have:

    x4 + 3x − 18 = (x2 + Ax + B)(x2 + Cx + D)

    Expanding the right hand side and equating coefficients, gives 4 equations in 4 unknowns.

    That gives somewhat of a road map.


    I found that the result of doing the above gave a cubic equation in A2 .
     
  16. Oct 3, 2016 #15

    epenguin

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    For the general quartic you always get this or a related cubic called 'the reducing cubic'whatever way you go about it.
     
    Last edited: Oct 3, 2016
  17. Oct 3, 2016 #16
    Lol guys and gals i had the class today and the teacher said she made a mistake. It was supposed to be x^4+3x^2-18 lol ,but thanks for the help and it was a stimulating and challenging thing for all us i think( maybe just me you guys are smarter than me.)
     
  18. Oct 3, 2016 #17
    substitute y = x^2 in for x and solve the quadratic in y; then take square root of y to get x
     
  19. Oct 4, 2016 #18

    epenguin

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    :approve: Yeah we told you she made that mistake :oldtongue: - but thanks for coming back to tell us. :oldsmile:
    It's practice - we don't always work things through ourselves to the answer, but we have had to get quite practiced in working out what the question was or should have been.
     
  20. Oct 4, 2016 #19
    Now I see, a bit late. Apparently the initial problem cannot be solved by substitution
     
  21. Oct 5, 2016 #20

    fresh_42

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    Quite the opposite. The "usual" mathematician is proud of not to know how to calculate other than ±1 * ±2. :smile:
     
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