Prove $\sum_{i=0}^n \binom{n}{i}^2x^{n-i} = 0$ has $n$ negative roots

Let's $n \in \mathbb{Z^+}$, how to $\text{prove}|\text{disprove}$ that:

the equation $\boxed{\sum_{i=0}^n \binom{n}{i}^2x^{n-i} = 0}$ has exactly $n$ distinct negative roots.

My friend get bored, then he started play with equations. And he found that $\sum_{i=0}^n \binom{n}{i}^2x^{n-i} = 0$ with $n = 1, 2, 3$ have $1, 2, 3$ distinct negative roots respectively, he asked me to prove the "quoted question" above; but seem like I can't do it, please help or give me a hint.

Thank you. Please help me edit my post & tags, I am not good at English

The polynomial in question is equal to $(1-x)^n P_n \left( \dfrac{1+x}{1-x} \right)$ where $P_n$ is the $n$-th Legendre polynomial. So the question becomes boils down to the fact that $P_n$ has $n$ distinct roots in $(-1,1),$ which is proved here.