Thread: polynomial root

If we are told that , where .

Prove that the polynomial has at least one real zero.

Originally Posted by putnam120

If we are told that , where .

Prove that the polynomial has at least one real zero.

As:



and:



as its continuous has a real root.

In fact all odd order polynomial with real coefficients must have at
least one real root.

RonL

Originally Posted by putnam120

If we are told that , where .

Prove that the polynomial has at least one real zero.

Use mean value theorem for integral.
Consider the continous function on , .

By the integral mean value theorem there is a number such as,

But that tells us that,

Evaluate,

Note only we know it has a solution we know it has a solution on the interval

Originally Posted by CaptainBlank

In fact all odd order polynomial with real coefficients must have at
least one real root.

But how do you know that

Originally Posted by ThePerfectHacker

But how do you know that

Well this partiular statement is independent of as I said any odd order
polynomial, but in this case yes I suppose could in principle
be zero.

In fact I do have another method using Descartes rule of signs which
does use the condition, which I beleive handles the possibility that
(but I don't intend to check that now).

RonL