Is there a rational univariat polynomial of degree 3 with 3 irrational roots?

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? en.wikipedia.org/wiki/Rational_root_theorem –  Micah May 27 '12 at 15:49
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 $n \ge 2$. –  Qiaochu Yuan May 27 '12 at 15:53
You're basically asking for a totally real field of degree 3 over $\mathbf{Q}$. See examples here. –  vgty6h7uij May 27 '12 at 15:55
@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. –  born May 27 '12 at 16:17
@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. –  born May 27 '12 at 16:53

1 Answer

The polynomial $f(x)=x^3-3x+1$ is an example for such a polynomial. To see this, check, that $f(-2)=-1$, $f(-1)=3$, $f(1)=-1$ and $f(2)=3$ which, by the intermediate value theorem, implies that $f$ has $3$ real roots. The rational root theorem now gives us a list of possible rational roots, namely $1$ and $-1$, which are both no roots. Therefore $f$ has three real, non-rational roots.

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