prove a polynomial has at least 2n-1 distinct real roots 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!
 A: 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.

 Consider a pair of consecutive roots $a<b$ of $P$ in $(1,\infty)$.  Let me first assume that $a$ and $b$ are both simple roots.  Then $P(a)=P(b)=0$ but $P'(a)$ and $P'(b)$ are nonzero with opposite sign.  It follows that there must be some $c\in(a,b)$ such that $P'(c)=-cP(c)$.  Note that $P(c)\neq 0$ since $a$ and $b$ are consecutive roots.  Similarly, there must exist $d\in (a,b)$ such that $P(d)=-dP'(d)$ and $P(d)\neq 0$.

 If $a$ or $b$ is a multiple root of $P$, the story is more complicated but the conclusion is the same.  Let us assume without loss of generality that $P$ is positive on $(a,b)$.  If $k$ is the multiplicity of the zero of $P$ at $b$, this means $P^{(j)}(b)=0$ for $j<k$ and $P^{(k)}(b)$ has the same sign as $(-1)^k$.  Now observe that the first nonzero derivative of $P(x)+xP'(x)$ at $b$ is the $(k-1)$st derivative which is $bP^{(k)}(b)$ (all the other terms involve lower derivatives of $P$ at $b$ which all vanish).  Thus for $x$ slightly less than $b$, $P(x)+xP'(x)$ has the same sign as $(-1)^{k-1}bP^{(k)}(b)$ which is negative.  But we can do a similar analysis at $a$ to conclude that $P(x)+xP'(x)$ is positive for $x$ slightly greater than $a$.  Thus $P(x)+xP'(x)$ must have a root in $(a,b)$.  We can again do a similar analysis for $P'(x)+xP(x)$ to show that it also has a root in $(a,b)$.

  So, the upshot is that $Q$ must have at least two distinct roots between each pair of roots of $P$ in $(1,\infty)$.  This gives $2n-2$ distinct roots of $Q$.  To find one more, let $a$ be the least root of $P$ in $(1,\infty)$ (if $n=0$ there is nothing to prove).  Note that $Q(a)\geq 0$.  Note also that if the leading term of $P$ is $bx^k$, then the leading term of $Q$ is $$x^2\cdot bx^k\cdot kbx^{k-1}+x\cdot b^2x^{2k}=(k+1)b^2x^{2k+1}.$$  In particular, $Q$ has odd degree and positive leading coefficient.   Thus $Q(x)\to-\infty$ as $x\to-\infty$.  It follows that $Q$ must have a root which is less than or equal to $a$.  Thus in total we find that $Q$ has at least $2n-1$ real roots, as desired.

