Roots of $p(x)+xp'(x)$ Let $p(x)$ be a polynomial of degree $n$ such that it has no real root (that is, $n$ is an even positive number).
Can we say anything about the roots of the polynomial $p(x)+xp'(x)$?
Here $p'(x)$ denotes the derivative of $p(x)$ with respect to $x$.
 A: Let's say that $p(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n$
Then we have $p'(x)=a_1+2a_2x+\cdots+na_nx^{n-1}$
And thus $xp'(x)=a_1x+2a_2x^2+\cdots+na_nx^n$
Now we want to find the roots of \begin{align}p(x)+xp'(x) &= (a_0+a_1x+a_2x^2+\cdots+a_nx^n)+ (a_1x+2a_2x^2+\cdots+na_nx^n)\\
&=a_0+(a_1+a_1)x+(a_2+2a_2)x^2+\cdots+(a_n+na_n)x^n\\
&=a_0+2a_1x+3a_2x^2+\cdots+(n+1)a_nx^n\\
&=\sum_{i=0}^n(i+1)a_ix^i\end{align}
A: Note that if $a$ is a root of $p$ with multiplicity strictly greater than $1$ (i.e
$$p(x) = (x-a)^k q(x)$$ for some $k > 1$ and $q$ a polynomial), then $a$ is a root of $p(x) + xp'(x)$. This follows from differentiating the above expression for $p$. When $a=0$ the above holds and is also true for $k=1$.
A: $$(xp(x))'=p(x)+xp'(x)$$
Now since $n$ (degree of $p(x)$) is even, we have $xp(x)$ an odd degree polynomial. Also, we can say that $xp(x)$ isn't monotonously increasing, this is so because a polynomial $q(x)$ of degree $k$ is is monotonously increasing, only if $$q(x)=a(x+b)^k+c \, ; a \neq 0$$ If we compare this with $xp(x)$, we'll get $c=b=0$ 
This suggests $p(x)=x^n$ which isn't possible since it has no real roots.
Now, since $xp(x)$ is an odd degree polynomial, it tend to $\pm \infty$ when $x \to \pm \infty$, and we have also proved that it isn't monotonous, thus it turns an even number of times, and whenever it turns, it's derivative will be $0$.
Hence we can conclude that the polynomial $p(x)+xp'(x)$ will have an even number of roots, with minimum $2$ roots.
