Sum polynomial and derivative How to prove that if polynomial $W(x)$ has $n$ real roots then
$\forall a \in \mathbb{R}$
$a W(x)+W'(x)$ has more than $n-1$ roots
I have no idea how to solve. Please some hint.
 A: First note that
$$f(x) = aW(x) + W'(x) = e^{-ax} \dfrac{d \left(e^{ax} W(x)\right)}{dx} \implies g(x) = e^{ax} f(x) = \dfrac{d \left(e^{ax} W(x)\right)}{dx}$$
Now any real root of $g(x)$ is also a real root of $f(x)$ and vice-versa.
The main idea to be used is between any two zeros of a smooth function, there is a zero for its derivative.
Consider a smooth function $h(x)$ with $n$ real roots. Let us assume that these roots are distinct. This means the function crosses the $X$ axis at $n$ distinct points and more importantly changes sign at these $n$ distinct points. Now between two roots there is always a local extremum since the function has to "bend" and come back to zero. This means the derivative is zero at some intermediate point. Hence, there is root of the derivative between two roots of the function. Once you see this, it is now clear that if we have $n$ roots for a function, then its derivative must have $n-1$ roots, since between any two roots of the function, there exists a point whose derivative is zero and hence this point is a zero for the derivative of the function. If you have a root of multiplicity $m$ at a point ,then this is also a root of its derivative with multiplicity $m-1$ and hence the same argument will work.
A: Posted an incorrect solution first, but I want to write @Marvis's answer in another way (so full credit goes to him):
Let $f(x) = e^{ax} W(x)$, then $f'(x) = e^{ax} (a W(x) + W'(x))$
$f(x)$ has exactly $n$ real roots and by mean value theorem between each of the roots, $f'(x)$ must have $n-1$ real roots.  But this means that $a W(x) + W'(x)$ has $n-1$ real roots.
A: Let $r_1<\ldots<r_l$ denote these $n$ real roots without multiplicity in increasing order, and let $m_k$ denote their multiplicity. So $n=m_1+\ldots+m_l$.
Every root of $W$ of multiplicity $m\geq 2$ is a root of $aW+W'$ with multiplicity $m-1$. This yields already $(m_1-1)+\ldots+(m_l-1)=n-l$ roots for $aW'+W$, counted with multiplicities.
Now
$$
aW(x)+W'(x)=W(x)\left( a+\frac{W'(x)}{W(x)}\right)=W(x)\left(a+\sum_{k=1}^l\frac{m_k}{x-r_k}\right)=W(x)F(x).
$$
On each interval $(r_k,r_{k+1})$, the rational function $F(x)$ is continuous with  $\lim_{r_k^+}F=+\infty$ and $\lim_{r_{k+1}^-}F=-\infty$. By the intermediate value theorem, there exists $x_k$ in this interval such that $F(x_k)=0$. Hence $(aW+W')(x_k)=0$. This yields as many extra roots as such intervals, namely $l-1$.
So we have exhibited $n-l+l-1=n-1$ real roots for $aW+W'$.
