Relation between the roots of $x^2+x+1$ and its derivative If $f(x)$ is a polynomial in n degree and has  $n$ real roots then is it necessary that $f'(x)$ has to have $n-1$ real roots? If this is so then $x^2+x+1$ has no real roots but the derivative of the polynomial has a root. Is it correct to conclude this?
 A: The derivative has a root because deg=1. No problemo because the implication goes in the other direction.
A: It is a necessary but not sufficient condition. You are challenging the "sufficient" property, not the "necessary" one.
A: If a polynomial $f(x)$ has $n$ real roots, then we can look at the (real) graph of $f(x)$. Now when a polynomial is zero you can have two situations:


*

*The polynomial intersects the $x$-axis

*The polynomial merely touches the $x$-axis


In the first case we can say that if our polynomial should have another root, than is has to "travel back to the $x$-axis". In other words: it has to undergo a change in its (vertical) direction, wich is only possible if the derivative is zero somewere.
So the important thing is that you now realise that between two consecutive roots, there is necessarily  some sort of turning point/horizontal tangent/root of the derivative. 
If you think this through a little you will be able to derive your statement.

If you look at case two  your statement need not be true. However the fact that we have $n=\deg(f)$ ensures us this won't happen.
