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?
Is there a rational univariat polynomial of degree 3 with 3 irrational roots?
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abstract-algebra
polynomials
irrational-numbers
rational-numbers
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2@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
1 Answers
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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.