TITLE: Proving all Solutions of a Polynomial Cannot all be Real
QUESTION [7 upvotes]: 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.

REPLY [11 votes]: 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.