MVT: Let $P(x)$ is a polynomial of $n$ degree and $P(x)=0$ has $n$ roots. Prove that $P(x)+P'(x)=0$ has at least $n$ roots. Let $P(x)$ is a polynomial of $n$ degree and $P(x)=0$ has $n$ roots. Prove that $P(x)+P'(x)=0$ has at least $n$ roots.
My idea:
Let $g(x)=e^{x}P(x)$ and $g'(x)=e^{x}(P(x)+P'(x))$.
Because $g(x)=0$ has $n$ roots so $g'(x)=0$ has at least $n-1$ roots, which means $P(x)+P'(x)=0$ has at least $n-1$ roots . However the problem need $P(x)+P'(x)=0$ has at least $n$ roots.
Please correct me where I was wrong.

From the problem above, I think the problem I mention below has some connections:
Let $ax^{2}+bx+c=0, a\neq0$ has two real roots, $P(x)$ is a polynomial with real coefficient and has $3$ real roots. Prove that $T(x)= aP(x)+bP'(x)+cP''(x)=0$ has at least $3$ real roots.
I do not know why we need the equation: $ax^{2}+bx+c=0$, so I do not have any ideas to prove the problem. Please help me!
Thank you so much!
 A: $Q=P + P^\prime$ has at least $n-1$ roots and is of degree $n$ as $P$. Therefore is has $n$ roots as you can write
$$Q(x)=P(x) + P^\prime(x) = a(x-r_1) \dots (x-r_{n-1})q(x)$$ where $r_1, \dots r_{n-1}$ are the $n-1$ roots of $P + P^\prime$ and $q$ a polynomial of degree $1$. Therefore $Q$ has only real roots.
Let's prove that those are distinct.
By hypothesis $g$ has $n$ real roots $g_1 < \dots < g_n$. Therefore $Q$ has $n-1$ distinct roots in $(g_1,g_n)$. Now $\lim\limits_{x \to -\infty} g(x) = g(g_1)=0$. Hence the differentiable map $g$ attains its extremum at $g_{n+1} \in (-\infty, g_1)$. And we have $g^\prime(g_{n+1})=Q(g_{n+1})=0$. Finally proving that $Q$ has $n$ distinct real roots.
A: IF $P(x)=0$ is ral polynomial of degree $n$ and it has $n$ real roots,
then by Rolle's theorem:$\frac{d}{dx} (e^x P(x))= e^x[(P(x)+P'(x)]=0$ will have at least $n-1$ real roots. Since the degree of $P(x)+P'(x)$ remains $n$ so it will have $n$ real roots, namely the last unpaired root has to be real as the polynomial is real.
Note: the reality of the roots is most important here.
