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?

By assumption (a,b,c being different reals)
$$W(x)=(x-a)(x-b)(x-c)Q(x)$$
then
$$W'(x)=((x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b))Q(x)+(x-a)(x-b)(x-c)Q'(x) \\
=(3x^2-2(a+b+c)x+ab+bc+ac) Q(x)+(x^3-(a+b+c)x^2 + (ab+bc+ac)x -abc) Q'(x)$$
and
$$W''(x)=(2(x-c)+2(x-a)+2(x-b))Q(x)+(x-a)(x-b)(x-c)Q''(x)+2((x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)) Q'(x) \\
=(6x-2(a+b+c)) Q(x) + (x^3-(a+b+c)x^2 + (ab+bc+ac)x -abc) Q'' + 2 (3x^2-2(a+b+c)x+ab+bc+ac) Q'(x)$$
Since we do not know Q any root of $P(x)$ must be a root of the "prefactor" polynomials
$$ 3x^2-2(a+b+c)x+ab+bc+ac) = 0 \\
x^3-(a+b+c)x^2 + (ab+bc+ac)x -abc = 0\\
6x-2(a+b+c) = 0 $$
And here I ran out of time and motivation.. Also no guarantee that any of that is correct, but should suffice
