Roots of a polynomial with binomial coefficients

Question

Let $f(x) = \sum\limits_{k=0}^n \binom{n}{k}^2x^k$. Show that polynomial $f(x)$ has $n$ distinct real roots, for $n\ge2$. What I've tried:
I tried to prove it by induction. Let $P(n): f(x) = (x-\alpha_1)(x-\alpha_2)\dots(x-\alpha_n)$, where $\alpha_1,\alpha_2\dots\alpha_n$ are the distinct real roots of polynomial.
Suppose that $P(n)$ is true and show that $P(n+1)$ is also true. $P(n+1):f(x) = \sum\limits_{k=0}^{n+1} \binom{n+1}{k}^2x^k$
Now I written $\binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}$ and thus $$ f(x) = \sum\limits_{k=0}^{n+1}\left(\binom{n}{k} + \binom{n}{k-1}\right)^2x^k = \sum\limits_{k=0}^{n+1}\left(\binom{n}{k}^2 + \binom{n}{k-1}^2 + 2\binom{n}{k}\binom{n}{k-1}\right)x^k \\= \sum\limits_{k=0}^{n+1}\binom{n}{k}^2x^k + \sum\limits_{k=0}^{n+1}\binom{n}{k-1}^2x^k+2\sum\limits_{k=0}^{n+1}\binom{n}{k}\binom{n}{k-1}x^k.$$ But what can I do now? I've tried several times but last term of this expression seems difficult to factorize.

Answer 1

Mathematica gives $$f(x)= \sum\limits_{k=0}^n \binom{n}{k}^2x^k= (1-x)^n P_n\left(\frac{1+x}{1-x}\right)$$ Where $P_n(z)$ are the well known Legendre Polynomials or order $n$ with distinct real zeros $z_n$. So we will have roots $x_n$ of $f(x)=0$ as $$x_n=\frac{z_n-1}{z_n+1}.$$