Can a polynomial with degree (r+s-1) have (r+s) distinct roots?

Question

While reading about generalised eigenvectors, i came to a strange proof. It goes like
If p is a polynomial which has r distinct roots {λ1,λ2...,λr}, and q is a polynomial with s distinct roots {μ1,μ2...,μs}.
We can have two polynomials a(of degree s-1),b(of degree r-1) such that a(x)=p(x)^-1,when x={μ1,μ2...,μs} and b(x)=q(x)^-1,when x={λ1,λ2...,λr}.
Now a(x).p(x)+b(x).q(x)=1 for x={μ1,μ2...,μs,λ1,λ2...,λr}, all being distinct by definition. Writing ap+bq-1=0, has r+s roots, while its a polynomial with degree at most (r+s-1). How does it works?
Reference http://math.mit.edu/~dav/generalized.pdf

Answer 1

Any nontrivial polynomial of degree at most $r+s-1$ can have at most $r+s-1$ distinct roots. So if $a(x)p(x) + b(x)q(x) - 1$ is equal to $0$ at $r+s$ points, that means it can only be the trivial polynomial which is $0$ everywhere.