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Factorize polynomial

  1. Sep 22, 2015 #1
    I should factorize following polynomial:

    P(x)=x^2n + 2cos(naπ)x^n + 1 in ℝ if i know that a is irrational number.

    Things that confuse me here are following:

    1. When factorizing polynomials, i have known exponents (unlike here, where i have 2n and n) so i don't know what to do with them?

    2. Why does it makes a difference if a is irrational number?
     
  2. jcsd
  3. Sep 22, 2015 #2

    jedishrfu

    Staff: Mentor

    Have you considered a substitution of variable x to make the equation a simple quadratic?
     
  4. Sep 22, 2015 #3
    You mean like: y=xn, but what can i do next, that cosine there is confusing, and fact that a is irrational makes it even more complicated.
     
  5. Sep 22, 2015 #4

    SteamKing

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    The value of the cosine in your polynomial doesn't depend on x, so it may be treated like a constant. Call it A and re-write your original polynomial:

    A = cos(naπ)

    P(x) = x2n + 2Axn + 1

    Now, make the substitution y = xn
     
  6. Sep 23, 2015 #5

    Ok, so now i have the following:

    y2 + 2Ay + 1

    y1,2=[-2A+-sqrt(4A2 - 4)]/2 = -A +- sqrt(A2 - 1)

    since A=cos(naπ) then A2=cos2(naπ) and A2 - 1 = cos2(naπ) - 1 =-sin2(naπ)

    but -sin2(naπ) could be anywhere between -1 and 1 so i don't know is it positive or negative, because if it is negative then i can't factorize it in ℝ.
     
  7. Sep 23, 2015 #6

    SteamKing

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    Obviously, in order to factor this polynomial in ℝ, certain conditions must be met; IOW, A2 - 1 ≥ 0, or A2 ≥ 1 or cos2(naπ) ≥ 1.

    Since -1 ≤ cos (θ) ≤ 1, then the polynomial can be factored in ℝ only if naπ = 2kπ, where k = 0, 1, 2, ...

    It's not clear why you would want to factor such a beast anyway. If you want to find the roots to P(x), it would seem that a numerical approach would work just as well.
     
  8. Sep 23, 2015 #7
    Well, it's just an exam question, anyway, what's with a, i mean why is it pointed out that a has to be irrational number, the way we solved this, it wouldn't be any different even if a wasn't irrational number?
     
  9. Sep 23, 2015 #8

    SteamKing

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    It's not clear what n is here. Is n any integer, or is it something else?

    Since a is irrational, this suggests that the condition naπ = 2kπ can be satisfied only if na = 0, and this will depend on what values n can take.
     
  10. Sep 23, 2015 #9
    I don't know about n, it's not defined what n is, but it might be integer (probably it is), anyway, since it is not defined i can't be sure what it is.
     
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