For what $a$ and $b$ does the polynomial $P(x)=(a+b) x^{5}+a b x^{2}+1$ divide by $x^{2}-3 x+2$?

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$x^{2}-3 x+2=(x-1)(x-2)$. Therefore, the equalities $P(1)=P(2)=0$ should hold, that is, $(a+b)+a b+1=(a+1)(b+1)=0,32(a+b)+4 a b+1=0$. From the first equality, we see that one of the numbers $a, b$ equals -1. Substituting into the second equality, we find that the second number equals ${ }^{31} / 28$.

## Otvet

$\{a, b\}=\{-1,31 / 28\}$.