Re: polynom+real zeros

On 29-04-2006 13:42, eugene wrote:
Prove that polynomial p(x)=x^100+x^98+a_3x^97+...+a_n cannot have
all real zeros.

I have got the following: if we suppose the contrary, we will have
that p'(x) must have at least 99 real zeros(between every pair of
consecutive real zeros of p there must be a real zerof of
p' --- (*) ), p''(x) must have at least 98 real zeros,... and so
on, p^(98)(x) must have at least two real zeros, but
p^(98)(x)=100!/2*x^2+1 wich surely cannot have real zeros.

Is it ok? The point where i'm a bit unsure is (*).

No, it is no ok. What you did prove was that p(x) cannot have 100 real
zeros. But, for instance, the polynomial x^100 only has one real zero
and it is still true that all of its zeros are real.

Best regards,

Jose Carlos Santos
.

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