= Solution
The <monic orthogonal polynomial> $p_n$ is orthogonal to every <polynomial> of degree less than $n$, since $p_0,\ldots,p_{n-1}$ form a basis of that polynomial space. Let $r_1,\ldots,r_k$ be its distinct interior roots at which its sign changes, and put $q(x)=\prod_{j=1}^k(x-r_j)$, with $q=1$ if $k=0$. The real polynomial $p_nq$ has constant sign throughout $(a,b)$: multiplication by each linear factor removes its corresponding sign change, while even-multiplicity roots do not change sign. It is not identically zero. Since $w>0$, its weighted integral is nonzero.
If $k<n$, orthogonality instead gives $\int_a^b p_nq\,w\,dx=0$, a contradiction. Thus $k\ge n$. Degree $n$ also gives $k\le n$, so $k=n$ and all roots have multiplicity one, with no roots left outside the interval. Therefore \b[there are exactly $n$ distinct simple zeros in $(a,b)$]. The weighted inner products are understood to exist, as required by the stipulated orthogonality.
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