Question

Question #672ae

Answer 1

See below.

Explanation:

Assuming

`p_3(x) = ax^3+bx^2+cx+d in P_3` we have that
if `p_3(x) in P^e` then `p_3(x) = p_3(-x)` or

`a x^3+b x^2+c x+d = a(-x)^3+b(-x)^2+c(-x)+d` or

`a x^3+b x^2+c x+d = -a x^3+b x^2-c x+d = 0, forall x` which implies on `a = c=0` hence

`p(x) = bx^2+d` and a basis for `W` is given by `{1, x^2}` with dimension `2`