In your conditions the result is not true, just consider
If n is odd or n even and , , .Thus there exists such that
, and .
Now you only need Bolzano.
Hi So I have to prove the following:
If with and and n is an even positive integer, then prove that p has at least two distinct roots.
So my proof looks a little like this so far:
p is continuous on .
Hence such that:
At this point Im stuck. I found, via graphing example polynomials, that the negative values cause the polynomial to have two distinct roots, but how to state this mathematically to finish the proof?
Any suggestions would be appreciated thanks.
Again, you don't prove it. It's not true. For example, has roots x= 0, x= 1, x= -1, three distinct roots, so it would certainly also be correct to say it has 2 distinct roots.
If you mean "does not have exactly two distinct roots" it is still not true. has exactly x= 0 and x= 1 as distinct roots.
What is true is that a polynomial, of odd degree, with real coefficients cannot have exactly two (or any even number) real roots (distinct or not). But that is not anything like what you have been saying.