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If a quartic has rational coefficients and one real root, how would one go about showing that the real root is rational?

I understand that the condition is equivalent to showing that having a polynomial with one irrational double root and two imaginary roots or one irrational quadruple root is impossible. The latter seems straightforward, since the ratio of the second coefficient to the first (which is rational) is the sum of the roots (which is irrational). Is there a "nice" way of showing the former?

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If $r$ is a double root and the other two roots are complex, i.e. $p(x) = (x-r)^2 (x - w) (x - \overline{w})$, then $x - r = \gcd(p(x), p'(x))$ which has rational coefficients.

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  • $\begingroup$ Thank you so much - what a beautiful solution! My calculus students enjoyed this immensely. I'm currently working on polynomials with my pre-calculus class (we've just finished Vieta's Formulas) and only a few know what a derivative is. Is there a non-calculus way of looking at it? $\endgroup$ – Robert Fitzgerald Nov 20 '14 at 2:10

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