How prove : polynomial $P(x)=W^{\prime \prime}(x) +(W^\prime(x))^2$ have a real root. Let $W(x)$ be a polynomial of degree> 2 having at least three different real roots. How prove : polynomial $P(x)=W^{\prime \prime}(x) +(W^\prime(x))^2$ have a real root?



Whether the assumption of the theorem can be weakened?
Consider the function $\phi\colon\mathbb{R}\to\mathbb{R}$, defined by $\phi(x)=e^{W(x)}$, and calculate
$$
\phi'=W'e^W, \qquad \phi''=(W''+(W')^2)e^W=Pe^W.
$$
Thus we have to prove that $\phi''(d)=0$ for some $d\in\mathbb{R}$.

We know that $\exists a&lt;b&lt;c\in\mathbb{R}$ such that $\phi(a)=\phi(b)=\phi(c)=1$.  Rolle's theorem gives that there exist $a_1\in[a,b]$ and $b_1\in[b,c]$ such that $\phi'(a_1)=\phi'(b_1)=0$. Applying the theorem again to $\phi'$ we get what we need.
Hint: Suppose that $W(x)$ has at least 3 different real roots called $a,$ $b$ and $c$ so $W(x)= (x-a)(x-b)(x-c) g(x)$ where $\deg(g)=\deg(W)-3$ and $g(a),$ $g(b)$ and $g(c)$ is non zeros. Then, you can compute find the expression of $P(x)$ in terms of $(x-a),$ $(x-b)$ and $(x-c).$