Let L be the operator on $P_3$ defined by
$L(p(x)=xp'(x)+p''(x)$
If $p(x)=a_0+a_{1}x+a_2(1+x^2), \text{calculate }L^{n}(p(x)).$
$p(x) = a_0+a_1x+a_2(1+x^2) \\ p'(x)=a_1+a_2(2x) \\p''(x)=2a_2 \\ L(p(x))=xp{'}(x)+p''(x)=a_1x+2a_2(1+x^2) \\ L^2(p(x))= L(L(P(x))=L(a_1x+2a_2(1+x^2))$
How does one derive the meaning of $L^{2}$. How how does one find $L^{n}$.
Another way would be induction. For $n=1$ we have $$\operatorname L(a_0+a_1x+a_2(1+x^2))=a_1x+2a_2(1+x^2).$$ Assume we have $$\operatorname L^n(a_0+a_1x+a_2(1+x^2))=a_1x+2^{n}a_2(1+x^2)$$ for a number $n\in \mathbf N$, then $$ \operatorname L^{n+1}(a_0+a_1x+a_2(1+x^2)) = \operatorname L(\operatorname L^n(a_0+a_1x+a_2(1+x^2)) \stackrel{\text{IH}}{=} \operatorname L(a_1x+2^{n}a_2(1+x^2)) =a_1x+2^{n+1}a_2(1+x^2).$$
Clearly, $L$ is a linear operator of $\mathbf P_3.$ Observe also that $$ L(1)=0, L(x)=x, \text{ and } L(1+x^2)=2(1+x^2), $$ that is, $L$ is diagonalizable in a basis $(1,x,1+x^2).$ It follows that $$ L^n(1)=0, L^n(x)=x, \text{ and } L^n(1+x^2)=2^n (1+x^2) $$ for all $n \ge 1.$ Therefore $$ L^n( a_0 +a_1x+a_2 (1+x^2))=a_1 x + 2^n a_2 (1+x^2)=2^n a_2x^2 +a_1 x+2^n a_2. $$ for all $n \ge 1.$
IMHO, the simplest way is to obtain the matrix $M$ of $L$ with respect to the basis $\{1,x,1+x^2\}$
Here is how.
$L(1)=0=0+0x+0(1+x^2)$, thus associated with coefficients $(0,0,0)$ for the first column of $M$.
$L(x)=x=0+1x+0(1+x^2)$, thus associated with coefficients $(0,1,0)$ for the second column of $M$.
$L((1+x^2)=2(1+x^2)=0+0x+2(1+x^2)$, thus associated with coefficients $(0,0,2)$ for the third column of $M$.
Thus
$$\tag{1}M=\pmatrix{0&0&0\\0&1&0\\0&0&2} \ \ \iff \ \ \pmatrix{0&0&0\\0&1&0\\0&0&2}\pmatrix{a_0\\a_1\\a_2}=\pmatrix{0\\a_1\\2a_2}$$
Remark: the second form of (1) could have been obtained directly.
As a consequence of (1), $$M^n=\pmatrix{0&0&0\\0&1&0\\0&0&2^n} \ \ \iff \ \ \pmatrix{0&0&0\\0&1&0\\0&0&2^n}\pmatrix{a_0\\a_1\\a_2}=\pmatrix{0\\a_1\\2^na_2}$$
which is equivalent to say that:
$$L^n(a_0+a_1x+a_2(1+x^2))=a_1x+2^{n}a_2(1+x^2)$$