Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 3 / 17B / a / Solution

If the monic orthogonal polynomial $p_n$ had fewer than $n$ distinct sign-changing zeros in $(-1,1)$, multiply the factors corresponding to those zeros to form a polynomial $q$ of degree below $n$. Then $p_{n}q$ has one sign and is not identically zero, so its positive-weight integral cannot vanish, contradicting orthogonality. Thus all $n$ zeros are distinct and lie in the open interval.