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If, a, b, c, d and e are all real numbers how could I prove that the 5 solutions of the equation: $$f(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e == 0$$ cannot all be real valued if: $$2a^2 < 5b$$

Any assistance is appreciated.

Nate222
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    Do you know about the factorization of polynomials into linear factors using their roots? – joriki Apr 07 '11 at 00:32

1 Answers1

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If there are 5 zeros, then the first derivative has 4 zeros, the second has 3 zeros, and the third has 2 zeros, all counted with multiplicity. The condition on $a$ and $b$ is exactly the condition that the third derivative has no real zeros.

lhf
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  • Thanks for the quick reply. I'm just a little confused still regarding how $2a^2 < 5b$ fits in with the third derivative of $f(x)$. Is it simply the fact that because $a^2$ is less than $b$ that there can't be any real zeros? – Nate222 Apr 07 '11 at 02:41
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    @NateyG, compute the discriminant of the third derivative. – lhf Apr 07 '11 at 02:49
  • Ah, I see. Got it now. Thanks. – Nate222 Apr 07 '11 at 04:09