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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.

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Do you know about the factorization of polynomials into linear factors using their roots? – joriki Apr 7 '11 at 0:32
up vote 11 down vote accepted

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.

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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 7 '11 at 2:41
@NateyG, compute the discriminant of the third derivative. – lhf Apr 7 '11 at 2:49
Ah, I see. Got it now. Thanks. – Nate222 Apr 7 '11 at 4:09

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