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The title pretty much asks my question: Does $f\in\mathbb{Q}[x]$ such that $$ f(x)=(x-\alpha_1)(x-\alpha_2)(x-\alpha_3),$$ where $\alpha_1, \alpha_2, \alpha_3\in\mathbb{R}\setminus\mathbb{Q}\ $ exist?

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    Do you know the rational root theorem? http://en.wikipedia.org/wiki/Rational_root_theorem2012-05-27
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    Yes, lots. Probably the simplest example is $f(x) = x^3 + x + 1$. More generally, the rational root theorem shows that $f(x) = x^n + x + 1$ has no rational roots for all $n \ge 2$.2012-05-27
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    You're basically asking for a totally real field of degree 3 over $\mathbf{Q}$. See examples [here](http://en.wikipedia.org/wiki/Cubic_field).2012-05-27
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    @Micah yes I know that theorem, but it only states necessary conditions on rational roots, right? But I'm asking for irrational roots.2012-05-27
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    @QiaochuYuan the polynomial $f(x)=x^3+x+1$ has two complex roots. So it is not an example for a polynomial with 3 irrational (but real) roots. [link](http://www.wolframalpha.com/input/?i=x^3%2Bx%2B1)2012-05-27
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    @vgty6h7uij: That's a good hint, I'll look at the two examples $x^3+x^2-2x-1\ $ and $x^3+x^2-3x-1\ $ and check whether none of the three real roots is rational.2012-05-27
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    @born: ah, I didn't see that you wanted the roots to be real. Well, then take $x^3 - 3x + 1$.2012-05-27
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    @born Micah's hint is fine. The rational root theorem gives a list of all possible rational roots. So take for example Qiaochu's last polynomial, check with some basic calculus that all the roots are real, and use the rational root theorem to conclude that all the possibilities for rational roots fail.2012-05-27
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    @RagibZaman yes, that's absolutely right and it works out perfectly. Thank you all! I'll post the answer to this question in around 6 hours, when the software here allows me to.2012-05-27

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