Multiplication by $x^2$ linear maps $T:P(\mathbb{R}) \mapsto P(\mathbb{R})$ defined by
$(Tp)(x) = x^2p(x)$
Verify that multiplication by $x^2$ is a linear map.
Additivity: $x^2(p+q) = x^2p+x^2q$
Homogeneity: $x^2(ap) = a(x^2p)$
Is this a correct verification?

It seems that you've just written down the definition of linearity, but I expect that the problem is looking for a more thorough proof using the definition of $P(\mathbb{R})$ - so it's really asking why the two equalities you wrote are true.
Let's start with polynomials $f = a_0 + a_1 x + a_2 x^2 + \cdots+ a_n x^n$, and $g = b_0 + b_1 x + \cdots + b_n x^n$ (we don't require $b_n \ne 0$ or $a_n \ne 0$). Then
\begin{align*}
x^2 (f + g) &= x^2 \Big((a_0 + b_0) + (a_1 + b_1) x + \cdots + (a_n + b_n)x^2\Big) \\
&= (a_0 + b_0) x^2 + (a_1 + b_1) x^3 + \cdots + (a_n + b_n) x^{n + 2} \\
&= \Big(a_0 x^2 + a_1x^3 + \cdots + a_n x^{n + 2}\Big) + \Big(b_0 x^2 + b_1 x^3 + \cdots+ b_n x^{n + 2}\Big) \\
&= x^2 (a_0 + \cdots + a_n x^n) + x^2 (b_0 + \cdots + b_n x^n) \\
&= x^2 f + x^2 g
\end{align*}
as desired. Now do something similar for multiplication by a scalar.
