= Solution
Let $x_1,\ldots,x_5$ be algebraically independent indeterminates and set $a_i=e_i(x_1,\ldots,x_5)$, the elementary symmetric polynomials. Then
$$
\prod_{i=1}^5(T-x_i)
=T^5-a_1T^4+a_2T^3-a_3T^2+a_4T-a_5.
$$
The <Fundamental theorem of symmetric polynomials> identifies the fixed field of the natural $S_5$ action on $\mathbb Q(x_1,\ldots,x_5)$ with $\mathbb Q(a_1,\ldots,a_5)=K$. Every permutation of the roots therefore gives a $K$-automorphism, and the <Galois group of a polynomial> is
$$
\operatorname{Gal}(f/K)\cong S_5.
$$
A characteristic-zero polynomial is <solvability by radicals>[solvable by radicals] exactly when its Galois group is a <solvable group>. Part (c) shows that $S_5$ is not solvable, so
$$
\boxed{f\text{ is not solvable by radicals over }K}.
$$
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