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?
Answer
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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