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For example: $$ x^3-6x^2+3x-10 $$

The rational roots test tells me that possible roots are $\pm\ 10, 5, 2, 1$. However, none of these roots will divide the polynomial into a more workable nominal.

How can I efficiently determine how to factor this without resources such as Wolfram|Alpha?

Thank you.

  • 1
    A cubic polynomial can only factor into the product of a linear and a quadratic factor (which itself might factor further). So if it has no linear factors, it's already irreducible.2012-08-12
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    @QiaochuYuan It has a linear factor; it just doesn't have a *rational* root.2012-08-12
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    I am assuming (based on the fact that the OP said "possible roots" and not "possible rational roots") that the OP wants to factor over $\mathbb{Q}$.2012-08-12
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    @QiaochuYuan You are correct2012-08-12
  • 0
    @Sven In that case, none of the tools mentioned below will give you a factorisation over $\mathbb Q$, since you have proven yourself that **there is no rational root**2012-08-13
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    To rephrase some of the answers below, factoring over $\Bbb{R}$ amounts to finding real roots of the polynomial.2012-08-13
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    @JenniferDylan: No, it does not. $x^4+1$ has no real roots, but it does factor (over $\mathbf R$) into $(x^2+\sqrt 2 x+1)(x^2-\sqrt 2 x+1)$. It does if degree is at most three.2012-08-13
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    For that matter, $$ x^4 + 4 = (x^2 + 2 x + 2)(x^2 - 2 x + 2) $$2012-08-13
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    @tomasz Oh my bad, I meant factoring into linear factors.2012-08-13

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