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Math Help - Function is irreducible in the rationals.

  1. #1
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    Function is irreducible in the rationals.

    Is a=3x^3+2x+2 reducible in the rationals?

    So a is reducible iff we can write a=bc, for bc in the rationals (polynomials) and b,c \neq0. So deg(a)=3, therefore 0<deg(b)<deg(a) and 0<deg(c)<deg(a). So either deg(a)=2 or deg(a)=1, since deg(b)+deg(c)=deg(a)=3

    The next step I'm not too sure about, but I believe it is to show that a(x) \neq0 for any x in the rationals. And this leads to the conclusion that a is irreducible since we cannot write a=bc since either b or c is a unit.

    Any help will be appreciated, sorry that my notation is not entirely accurate in this post (not used to LateX).
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Function is irreducible in the rationals.

    Quote Originally Posted by A6363 View Post
    Is a=3x^3+2x+2 reducible in the rationals?

    So a is reducible iff we can write a=bc, for bc in the rationals (polynomials) and b,c \neq0. So deg(a)=3, therefore 0<deg(b)<deg(a) and 0<deg(c)<deg(a). So either deg(a)=2 or deg(a)=1, since deg(b)+deg(c)=deg(a)=3

    The next step I'm not too sure about, but I believe it is to show that a(x) \neq0 for any x in the rationals. And this leads to the conclusion that a is irreducible since we cannot write a=bc since either b or c is a unit.

    Any help will be appreciated, sorry that my notation is not entirely accurate in this post (not used to LateX).
    You've got a handle on things. If there are no rational roots to a(x) = 0, then it cannot be factored over the rationals. In fact, that's all you need to do. You don't need to worry about the degrees.

    And your LaTeX is good!

    -Dan
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    Re: Function is irreducible in the rationals.

    The "rational root theorem" will be very useful here: if the polynomial equation, a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0, has rational root x= \frac{m}{n}, where m and n are integers with no common factors, then the numerator, m, must evenly divide the constant term, a_0, and the denominator, n, must evenly divide the leading coefficient, a_n. Here, that means that the numerator must be one of 1, -1, 2, or -2 while the denominator must be one of 1, -1, 3, or -3. That gives 1, -1, 2, -2, 2/3, and -2/3 as the only possible rational roots. Try them and if none actually are roots of the equation, it has no rational roots and the polynomial is "irreducible over the rationals".
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    Re: Function is irreducible in the rationals.

    Thank you for your help.
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