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Thread: Irreduclible polynomials

  1. #1
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    Irreduclible polynomials

    How can I show that these polynomails are irreducible over rationals.


    f(x) = x^4+x-1
    g(x)= x^4+8x+12
    h(x) = 1+x+x^2/2+x^3/6+x^4/24
    j(x) = x^4-4x-1
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  2. #2
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    Quote Originally Posted by ZetaX View Post
    f(x) = x^4+x-1
    Write $\displaystyle x^4 + x - 1 = (x^2 + ax + b)(x^2+bx+d)$ where $\displaystyle a,b,c,d\in \mathbb{Z}$ and show this is impossible by comparing coefficients.

    You can try similar arguments for the others.
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  3. #3
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    Hello, ZetaX!

    I would assume you're expected to use the Rational Roots Theorem.

    Given a polynomial $\displaystyle p(x)$, if $\displaystyle p(a) = 0$, then $\displaystyle (x-a)$ is a factor of $\displaystyle p(x).$


    How can I show that these polynomials are irreducible over rationals?

    $\displaystyle (1)\;\;f(x) \:=\: x^4+x-1$

    $\displaystyle (2)\;\;g(x)\:=\: x^4+8x+12$

    $\displaystyle (3)\;\;h(x) \:=\: 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\;\; {\color{blue}=\;\frac{1}{24}\left(x^4 + 4x^3 + 12x^2 + 24x + 24\right)}$

    $\displaystyle (4)\;\;j(x) \:=\: x^4-4x-1$

    All the polynomials have a leading coefficient of 1.
    . . So the only possible rational roots are factors of the constant term.

    For (1), the only choices are: .$\displaystyle \pm1\quad\hdots $ Try them in $\displaystyle f(x).$

    For (2), the only choices are: .$\displaystyle \pm1, \pm2, \pm3, \pm4, \pm6, \pm12$

    For (3), the only choices are: .$\displaystyle \pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm12, \pm24$

    For (4), the only choices are: .$\displaystyle \pm1$

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, ZetaX!

    I would assume you're expected to use the Rational Roots Theorem.

    Given a polynomial $\displaystyle p(x)$, if $\displaystyle p(a) = 0$, then $\displaystyle (x-a)$ is a factor of $\displaystyle p(x).$



    All the polynomials have a leading coefficient of 1.
    . . So the only possible rational roots are factors of the constant term.

    For (1), the only choices are: .$\displaystyle \pm1\quad\hdots $ Try them in $\displaystyle f(x).$

    For (2), the only choices are: .$\displaystyle \pm1, \pm2, \pm3, \pm4, \pm6, \pm12$

    For (3), the only choices are: .$\displaystyle \pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm12, \pm24$

    For (4), the only choices are: .$\displaystyle \pm1$

    Be careful if a fourth degree polynomial has no zeros does not mean it is irreducible.
    Consider, $\displaystyle x^4+2x^2+1$, no rational zeros but it can be factored.
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