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Math Help - Rational Roots

  1. #1
    Senior Member sfspitfire23's Avatar
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    Rational Roots

    We're going over some rational root theorem stuff in class. Heres some Q's

    If I use the rational root theorem to find possible roots, but none of those roots exist, is the polynomial unable to be factored? So, if the polynomial has no roots it cant be factored?

    Another Q-

    Today we looked at factor rings of polynomials over a field. I know the theorem: p(x) is irreducible over F iff F[x]/\left< p(x) \right>.

    Now, say Im looking at a field order 25. I then consider F=\mathbb{Z}_5. Heres the question. Can i just guess and check polynomials to see if they're irreducible in Z5 and put that in for <p(x)>? or is there a certain method to this?
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    Quote Originally Posted by sfspitfire23 View Post
    We're going over some rational root theorem stuff in class. Heres some Q's

    If I use the rational root theorem to find possible roots, but none of those roots exist, is the polynomial unable to be factored? So, if the polynomial has no roots it cant be factored?


    Not at all: x^4+2x^2+1=(x^2+1)^2 has no rational roots, and still it can be reduced (or factored).
    From the non-existence of roots you can deduce irreducibility ONLY when the degree of the polynomial is \le 3


    Another Q-

    Today we looked at factor rings of polynomials over a field. I know the theorem: p(x) is irreducible over F iff F[x]/\left< p(x) \right>.

    ...if and only if....WHAT!??

    Now, say Im looking at a field order 25. I then consider F=\mathbb{Z}_5. Heres the question. Can i just guess and check polynomials to see if they're irreducible in Z5 and put that in for <p(x)>? or is there a certain method to this?
    You have to take an irreducible pol. of degree 2 over \mathbb{Z}_5, and yes: you can do this "by eye". i.e.: guessing. It's not that hard. Of course, you could first find out what elements in \mathbb{Z}_5 aren't squares and then build a quadratic whose discriminant is one of those non-squares and thus get an irreducible quadratic, but guessing is fine and pretty fast, too.

    Tonio
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    Quote Originally Posted by sfspitfire23 View Post
    We're going over some rational root theorem stuff in class. Heres some Q's

    If I use the rational root theorem to find possible roots, but none of those roots exist, is the polynomial unable to be factored? So, if the polynomial has no roots it cant be factored?
    It has no rational roots so whether it can be factored or not depends upon what you mean by "factored". If you mean "factored with integer or rational coeffcients", no it can't. But if you mean "factored with real or complex coefficients". By the "rational root theorem", any rational roots of x^2- 2= 0 or x^2+ 2= 0 must be integers dividing 2- so one of 1, -1, 2, or -2. Clearly none of those satisfy the equations so they have no rational roots and cannot be factored with integer or rational coefficients. But we can factor x^2- 2= (x- \sqrt{2})(x+\sqrt{2}) and x^2+ 2= (x-i\sqrt{2})(x+i\sqrt{2}).

    Another Q-

    Today we looked at factor rings of polynomials over a field. I know the theorem: p(x) is irreducible over F iff F[x]/\left< p(x) \right>.

    Now, say Im looking at a field order 25. I then consider F=\mathbb{Z}_5. Heres the question. Can i just guess and check polynomials to see if they're irreducible in Z5 and put that in for <p(x)>? or is there a certain method to this?
    Last edited by HallsofIvy; December 13th 2009 at 06:46 AM.
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