Prove the zeroes of a polynomial are all real and distinct 
For a polynomial $P(x) = (x-x_1)(x-x_2)\cdots (x-x_n)$ with distinct real zeroes, $x_1 < x_2<\cdots < x_n$, prove or disprove that all zeroes of $f(x) := P'(x) - kP(x)$ are real and that for any two distinct zeroes $y_0, y_1$ of $f(x)$, there is a zero of $P(x)$ between them.

I think it would be useful to apply Rolle's theorem to the function $e^{-kx} P(x)$, which has the same zeroes as $P(x)$. The derivative is $e^{-kx}f(x)$. The issue is that Rolle's theorem only ascertain $n-1$ real and distinct zeroes of $e^{-kx}f(x)$, while if $k\neq 0$, there may be another zero. How can I prove that this extra zero is real and that it is smaller than the smallest zero of $P(x)$ or larger than the largest zero of $P(x)$ (this would satisfy the requirement since the $n-1$ zeroes of $e^{-kx}P(x)$ guaranteed by Rolle's theorem are in-between two consecutive zeroes of $P(x)$).
 A: Assuming $n>1.$
$P$ and $P'$ have no common zeroes so $0=P'(x)-kP(x)\iff k=\frac {P'(x)}{P(x)}=\sum_{j=1}^n\frac {1}{x-x_j}.$
Let $1\le j<n.$ For each $i$ such that $j\ne i\ne j+1,$ the function $\frac {1}{x-x_i}$ is continuous, and hence bounded, on the interval $ [x_j,x_{j+1}],$ while the function $g_j(x)=\frac {1}{x-x_j}+\frac {1}{x-x_{j+1}}$ is continuous on $(x_j,x_{j+1})$ with $\lim_{x\to x_j^+}g_j(x)=+\infty$ and $\lim_{x\to x_{j+1}^-}g_j(x)=-\infty.$ Therefore $\{P'(x)/P(x):x\in (x_j,x_{j+1})\}=\Bbb R.$ So there exists $y_j\in (x_j,x_{j+1})$ with $P'(y_j)/P(y_j)=k.$
The polynomial $f(x)=P'(x)-kP(x)$ has degree $n$ or less and has real co-efficients and has $n-1$ zeroes $y_1,...,y_{n-1}$ so the $n$th zero of $f(x),$ if it exists, must be in $\Bbb R.$
Because if $z\in \Bbb C\setminus \Bbb R$ and $f(z)=0$ then $\bar z\ne z$ and $0=\overline {f(z)}=f(\bar z)$ (because $f$ has real co-efficients), but then $f$ is a polynomial of degree $n$ or less with at least $n+1$ zeroes $z,\bar z,y_1,...,y_{n-1},$  implying $f$ is identically $0.$ But $f(x_1)=\prod_{j=2}^n(x_1-x_j)\ne 0,$ a contradiction.
