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I want to prove that $x^7-120x^5+1875x^3+12500x+5$ has exactly 5 real roots. I have been able to show (by plugging in values and using the intermediate value theorem) that there are at least 5, but I don't know how to prove that there are exactly 5.

By the way, I am trying to prove this as a lemma to show that the polynomial has $S_7$ as its Galois group. See Is there an irreducible, degree 7 polynomial in $\mathbb{Q}[x]$ which has exactly two of its roots in $\mathbb{C-R}$?

Pascal's Wager
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  • If you can compute the discriminant and it is negative, then there must be (an odd number of) non-real complex-conjugate root pairs as well. – ccorn May 30 '18 at 20:34
  • That sounds awfully painful, though. – Pascal's Wager May 30 '18 at 20:35
  • @Pascal'sWager http://www.wolframalpha.com/input/?i=Discriminant%5Bx%5E7-120x%5E5%2B1875x%5E3%2B12500x%2B5%5D executed fairly quickly (and did return a negative number, albeit one with 36 decimal digits). – Daniel Schepler May 30 '18 at 21:30

1 Answers1

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There are two sign changes in the coefficients, so there are either 2 or 0 positive roots, by Descartes's rule of signs. Replacing $x$ by $-x$ gives $-x^7+()x^5-()x^3-()x+()$, where all the $()$s are positive, which has 3 sign changes, so there are either 3 or 1 negative roots. Hence there are at most 5 real roots.

Chappers
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