Final answer:
Subsets B and D are subspaces of P_2.
Step-by-step explanation:
Subset A: This subset is not a subspace of P_2 because the zero vector, which is the constant function 0, is not in the set. Therefore, it fails the subspace test for containing the zero vector.
Subset B: This subset is a subspace of P_2 because it satisfies all three conditions for being a subspace. The zero vector is in the set, the set is closed under addition and scalar multiplication, and it contains the additive inverse of any element in the set.
Subset C: This subset is not a subspace of P_2 because it fails the scalar multiplication test. The zero vector is not in the set and scalar multiplication does not keep an element in the set.
Subset D: This subset is a subspace of P_2 because it satisfies all three conditions for being a subspace.
Subset E: This subset is not a subspace of P_2 because it fails the scalar multiplication test. The zero vector is not in the set and scalar multiplication does not keep an element in the set.
Subset F: This subset is not a subspace of P_2 because it fails the scalar multiplication test. The zero vector is not in the set and scalar multiplication does not keep an element in the set.