 Solution

Over ${\mathbb{F}}_5$,The quadratic discriminant is $9-16\equiv 3$, while the squares modulo five are $0,1,4$. Thus that factor is irreducible. Adjoining either quadratic root creates a degree-two extension and supplies the other root by the sum-of-roots relation. Consequently the splitting field is ${\mathbb{F}}_{25}$ and $\boxed{[K:{\mathbb{F}}_5]=2}$.