Why does saying that $\mathscr{P}_2$ is isomorphic to $R^3$ immediately prove that a constant-wise addition of objects in $\mathscr{P}_2$ is an inner product?
What I mean by constant-wise addition is this operation:
let $p(x) = a+bx+cx^2$ and let $q(x) = d + ex + fx^2$. The operation in question is: $\langle p(x),q(x)\rangle = ad + be + cf $
To show that that operation is an inner product on $\mathscr{P}_2$, we need only show that the dot product in $R^3$ is an inner product, because $\mathscr{P}_2$ is isomorphic to $R^3$ (and the dot product is indeed an inner product on $R^2$).
Why is that? What significance does $\mathscr{P}_2$ being isomorphic to $R^3$ have in this context?
Let $\phi: \mathscr P_2 \to \Bbb R^3$ be the isomorphism of vector spaces given by $\phi(a + bx + cx^2) = (a,b,c)$. Let $\langle \cdot, \cdot \rangle_{\Bbb R^3} : \Bbb R^3 \times \Bbb R^3 \to \Bbb R$ denote the usual dot-product. Your "constant-wise addition" can be written as the map $( \cdot , \cdot): \mathscr P_2 \times \mathscr P_2 \to \Bbb R$ defined by $$ (p,q) = \langle \phi(p), \phi(q) \rangle_{\Bbb R^3} \qquad p,q \in \mathscr P_2. $$ Using the fact that $\phi$ is an isomorphism of vector spaces (an invertible linear transformation) and that $\langle \cdot, \cdot \rangle_{\Bbb R^3}$ is an inner-product, we can show that $( \cdot , \cdot)$ as defined above satisfies the definition of an inner product.