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Finding all zeros of a Polynomial

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Find all zeros of k(x) = 4x^4-7x^2+3x

    2. Relevant equations

    n/a

    3. The attempt at a solution

    I don't know if I should find the factors of 0 and 4.. Can you help me what to start out with.. I can do other ones like p(x) = 3x^3-37x^2+84x-24. But this problem k(x) = 4x^4-7x^2+3x doesn't have the last number to factor. can't factor the 0..
     
  2. jcsd
  3. Oct 11, 2011 #2

    micromass

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    Factor out x first.
     
  4. Oct 11, 2011 #3

    symbolipoint

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    k(x) = 4x^4-7x^2+3x

    Start with k(x) = x(4x3 - 7x + 3)

    Try to use Rational Roots Theorem to look for rational zeros. One of the zeros is obviously x=0, but you are now interested in the zeros of the degree-three polynomial.
     
  5. Oct 11, 2011 #4
    Oh. I didn't know you could factor it. I got these zeros. 0, +-1, +-3, +-3/2, +-3/4.
     
  6. Oct 11, 2011 #5

    gb7nash

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    They're not necessarily zeros (and it's impossible to have this many zeros since you're looking at a polynomial of degree 3). Now test each value and see if you obtain 0.
     
  7. Oct 11, 2011 #6
    Ok. I got x=0, x=1, and x=-3/2.
     
  8. Oct 11, 2011 #7

    symbolipoint

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    The function has one additional Real zero. All four zeros appear to be rational when checked quickly on a graphing calculator.
     
  9. Oct 11, 2011 #8
    I don't have a graphing calculator atm. Can you tell me which is the other one? I couldn't find it using the synthetic division.
     
  10. Oct 11, 2011 #9

    symbolipoint

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    You do not really need a graphing calculator; I just used one to check "quickly" what to expect. What you should do is recheck the Rational Roots Theorem, and make sure that you found all the roots which should be tested. You may have easily missed a couple or so of them.
     
  11. Oct 12, 2011 #10

    SteamKing

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    For a fourth order equation, if you have found three of the roots, you should be able to reduce the original equation to find the final remaining root. Recheck your division.
     
  12. Oct 12, 2011 #11

    NascentOxygen

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    There are plenty of web resources for plotting graphs. http://fooplot.com/index.php?q0=4x^4-7x^2+3x

    A graph will allow you to get visually close to the value, though you still need to check to see whether the value is an exact solution.
     
  13. Oct 12, 2011 #12

    HallsofIvy

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    Yes, you can immediately factor out x to get [itex]x(4x^3- 7x+ 3)[/itex]. As you say, the only possible rational roots of that cubic are 1, -1, 1/4, -1/4, 1/2, -1/2, 3/4, -3/4, 3/2, -3/2. You don't need a graphing calculator to see that [itex]4(1^3)- 7(1)+ 3= 0[/itex]!

    Dividing [itex]4x^3- 7x+ 3[/itex] by x- 1 you get [itex]4x^2+ 4x- 3[/itex]. So you have a [itex]4x^4- 7x^2+ 3x= x(x- 1)(4x^2+ 4x- 3)[/itex]. That last quadratic is easy to factor- although you could also use the quadratic formula if it weren't.
     
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