Question

Let P^(e)= {p(x)€R[x] | p(x)=p(-x) }
P^(o)={p(x)€R[x] | p(x)=-p(-x) }
Check that P^(e) and P^(o) are subspaces of R[x].

Answer 1

$$P^{(e)}= \{p(x) \in R[x] | p(x)=p(-x) \}$$

$$r,s \in P^{(e)} \Rightarrow (r+s)(x)=r(x)+s(x)=r(-x)+s(-x)=(r+s)(-x)$$

$$r \in P^{(e)}, a \in R \Rightarrow (ar)(x)=ar(x)=ar(-x)=(ar)(-x)$$

$$P^{(o)}= \{p(x) \in R[x] | p(x)=-p(-x) \}$$

$$r,s \in P^{(o)} \Rightarrow (r+s)(x)=r(x)+s(x)=-r(-x)-s(-x)=-(r+s)(-x)$$

$$r \in P^{(o)}, a \in R \Rightarrow (ar)(x)=ar(x)=-ar(-x)=-(ar)(-x)$$