OB. The set is a subspace of P₂. The set contains the zero vector of P₂, the set is closed under vector addition, and the set is closed under multiplication by scalars.
To show this, we need to verify the three conditions for a set to be a subspace:
The set contains the zero vector: The zero vector of P₂ is 0t² = 0, which is in the set since any real number multiplied by 0 is 0.
The set is closed under vector addition: Let p(t) = at² and q(t) = bt² be two polynomials in the set. Then p(t) + q(t) = (a +