Comparing real roots of $P(x)$ and $P'(x)$

Question

Let $a$ be a real number and $P(x)$ be a polynomial with real coefficients.

1) Prove that $P'(x)$ doesn't have more non real roots than $P(x).$

2) $aP(x)+P'(x)$ doesn't have more non real zeroes than the polynomial $P(x)$ itself.

I tried it like this.

$P'(x)$ has less degree than $P(x),$ so if it assumes more non real values than $P(x)$ then it assumes more real roots than $P'(x).$ Then it can be said that $P'(x)$ is a constant polynomial.

But i don't think my argument is right.

Hi, I changed your title so that it is more suggestive. Hope you do not mind. — , Apr 24 '14 at 04:04
basically the problem is that i don't know how to write in latex..@John — soumajit das, Apr 24 '14 at 04:07

score 1 · Answer 1 · answered Apr 24 '14 at 05:07

Hint to 1: Between any two real roots of $P$, there is at least one real root of $P'$. The applies even if the real roots of $P$ are repeated. Hence, what is the relationship between the number of real roots of $P$ and the number of real roots of $P'$?

Apply the fundamental theorem of algebra to relate it to non-real roots.

Comparing real roots of $P(x)$ and $P'(x)$

1 Answers1