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Is there an irreducible, degree 7 polynomial in $\mathbb{Q}[x]$ which has exactly two of its roots in $\mathbb{C-R}$? If so, what is it?

I'd prefer a quick, concrete example rather than an abstract proof.

Pascal's Wager
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    See https://math.stackexchange.com/questions/1194928/irreducible-polynomial-in-mathbb-qx-of-degree-n-having-exactly-n-2-real – lhf May 30 '18 at 19:40

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I think probably $p(x) = x(x^2+130)(x-5)(x+5)(x-10)(x+10) + 5 = x^7 + 5 x^5 - 13750 x^3 + 325000 x + 5$ should work. In a quick graph, it still appears to have exactly five real roots (as expected being a small perturbation from $x(x^2+130)(x-5)(x+5)(x-10)(x+10)$), and it was constructed to satisfy Eisenstein's criterion with $p = 5$.

To prove it has exactly five real roots: first, by substituting for example $x=-13, -8, -3, 2, 7, 12$ you should get alternating signs. Therefore, using the intermediate value theorem on each interval, you get at least five roots. Also, for example if you calculate the gcd of $p$ and $p'$, you should get 1, implying that none of these roots can be double roots. On the other hand, the fifth derivative $p^{(5)}(x) = 2520 x^2 + 600$ has no real roots, which implies that there can be at most five real roots of $p$. (The argument here: suppose $p$ had six distinct real roots. Then applying Rolle's Theorem on the intervals between these roots, you would get at least five distinct roots of $p'$; then at least four distinct roots of $p''$; and so on until you would get at least one root of $p^{(5)}$.)

Daniel Schepler
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  • I see by checking the values at particular points and using the intermediate value theorem, there are at least 5 real roots. But how can we show there are exactly 5? – Pascal's Wager May 30 '18 at 20:03
  • @Pascal'sWager I modified the example slightly to allow use of a trick which proves the modified polynomial has at most 5 real roots. – Daniel Schepler May 30 '18 at 20:46