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Math Help - x^3+2x^2-x+5

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    Post x^3+2x^2-x+5

    Prove that the following polynomials has no integer roots
    x^3+2x^2-x+5

    The polynomial has no roots mod 2.

    Thank you very much.
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  2. #2
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    Quote Originally Posted by Jenny20 View Post
    Prove that the following polynomials has no integer roots
    x^3+2x^2-x+5

    The polynomial has no roots mod 2.

    Thank you very much.
    In both cases the rational roots theorem works.

    The above polynomial will only have rational roots if those rational roots are of the form:
    \frac{\text{factors of 5}}{\text{factors of 1}}
    To be precise the numerator is factors of the constant term, the denominator is factors of the coefficient of the leading term.

    So in this case we only have four possible rational roots: \pm1, \pm5. Check these and you have your answer.

    The same works for the polynomial mod(2):
    x^3+2x^2-x+5 \equiv x^3 + x + 1 (mod 2)

    Here we only have two possibles: \pm 1.

    -Dan
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    Quote Originally Posted by Jenny20 View Post
    Prove that the following polynomials has no integer roots
    x^3+2x^2-x+5

    The polynomial has no roots mod 2.

    Thank you very much.
    A useful rule to know is that a 2nd or 3rd degree you can just substitute values into it to see if it is zero.
    Since we are mod 2 we are working in the ring \mathbb{Z}_2=\{0,1\}
    Upon evaluation none of these work.
    Thus it has no zero.
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    Every cubic with real coefficients has at least 1 real solution.

    The discriminant of a cubic is:

    4b^{3}d-b^{2}c^{2}+4ac^{3}-18abcd+27a^{2}d^{2}

    If the discriminat is >0, then there is one real root and the other 2 are complex conjugates.

    Checking this discriminant with a=1, b=2, c=-1, d=5, we get D=1007.

    Therefore, there is one real root. The other 2 are complex, so they are eliminated as integer roots.

    By the Intermediate Value Theorem, we find a sign change between -2 and -3, because f(-2)=7 and f(-3)=-1.

    The one real root is between -2 and -3, therefore, it can not be an integer.

    Actually, it's x=-2.925851551....

    Just for fun, here's the factorization:

    (x+2.925851551)(x-0.4629257757+1.222539948i) (x-0.4629257757-1.222539948i)
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    Quote Originally Posted by galactus View Post
    Every cubic with real coefficients has at least 1 real solution.
    Yes, in \mathbb{R}[x].
    Over here the question is being asked is in a different ring \mathbb{Z}_2[x][.
    Thus, the theorem does not apply.
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