If $P(x)$ is a real polynomial has $n$ distinct real roots in $(1,+\infty)$. Set:
$$Q(x)=(x^2+1)P(x)P'(x)+x(P^2(x)+P'^2(x))$$
How to prove $Q(x)=0$ has at least $2n-1$ distinct real roots?
I think the idea is finding closed intervals that the boundary points have different signs. I think the only special points are the roots of $P(x)$ or $P'(x)$. But if $a$ is such a point, we have $Q(a)>0$. So i'm not sure this can work out… Can anyone show me some hints? Thanks!
Best Answer
Dividing by $P(x)^2$, we get that assuming $P(x)\neq 0$, $Q(x)=0$ is equivalent to $$xR^2+(x^2+1)R+x=0$$ where $R=P'(x)/P(x)$. Considering this as a quadratic in $R$, the roots are $R=-x$ and $R=-1/x$. So, $Q(x)=0$ iff $$P'(x)=-xP(x)\text{ or }P(x)=-xP'(x)$$ (note that these conditions also include the possibility that $P(x)=0$, in which case we easily see that $Q(x)=0$ is equivalent to $P'(x)=0$ or $x=0$). Note that it is impossible for both $P'(x)=-xP(x)$ and $P(x)=-xP'(x)$ to hold at once for $x>1$ if $P(x)\neq 0$, since together they imply $x=1/x$ (this is where the assumption that the roots of $P$ are greater than $1$ will come in).
Now you can try and use this characterization to find roots of $Q$ between the roots of $P$, by considering how the signs of $P'(x)+xP(x)$ and $P(x)+xP'(x)$ change. The details are hidden below.