Solution
= 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 $p_nq$ 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.