Answer
See below
Work Step by Step
Given: $T_2T_1:P_4(R)\rightarrow R$
a) Computing $[T]^C_B$
$(T_2T_1)(1)=T_2(T_1(1))=T_2(1')=T_2(0)=0\\
(T_2T_1)(x)=T_2(T_1(1))=T_2(x'')=T_2(1)=1\\
(T_2T_1)(x^2)=T_2(T_1(x^2))=T_2((x^2)')=T_2(2x)=2.2=4\\
(T_2T_1)(x^3)=T_2(T_1(x^3))=T_2((x^3)')=T_2(3x^2)=3.2^2=12\\
(T_2T_1)(x^4)=T_2(T_1(x^4))=T_2((x^4)')=T_2(4x^3)=4.2^3=32$
then we have $(T_2T_1)(1)=0.1 \rightarrow [(T_2T_1)(1)]=_C=[0]\\
(T_2T_1)(x)=1.1 \rightarrow [(T_2T_1)(x)]=_C=[1]\\
(T_2T_1)(x^2)=4.1 \rightarrow [(T_2T_1)(x^2)]=_C=[4]\\
(T_2T_1)(x^3)=12.1 \rightarrow [(T_2T_1)(x^3)]=_C=[12]\\
(T_2T_1)(x^4)=32.1 \rightarrow [(T_2T_1)(x^4)]=_C=[32]$
Hence, $[T_2T_1]_A^C=\begin{bmatrix}
0 & 1 & 4 & 12 & 32\end{bmatrix}$
b) By Theorem 6.4.7
$$[T_2T_1]^C_A=[T_2]^C_B[T_1]^B_A$$
We can notice that $[T_1]_A^B=\begin{bmatrix}
0 & 1 & 0 & 0 & 0\\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0\\ 0 & 0 & 0 & 0 & 4 \end{bmatrix}$
$[T_2]^C_B=\begin{bmatrix}
1 & 2 & 4 & 8\end{bmatrix}$
Consequently, $[T(v)]_C=[T]^C_B[A]_B=\begin{bmatrix}
1 & 2 & 4 & 8\end{bmatrix}\begin{bmatrix}
0 & 1 & 0 & 0 & 0\\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0\\ 0 & 0 & 0 & 0 & 4 \end{bmatrix}
\rightarrow T(A)=\begin{bmatrix}
0 & 1 & 4 & 12 &32 \end{bmatrix}$
c) From Theorem 6.4.5
$$[(T_2T_1)(p(x))]_C=[T_2T_1]^C_A[p(x)]_A$$
We have $[p(x)]_A=\begin{bmatrix}
2 \\ 5\\ -1 \\0 \\ 3 \end{bmatrix}$
Obtain: $[(T_2T_1)(p(x))]_C=\begin{bmatrix}
2 \\ 5\\ -1 \\0 \\ 3 \end{bmatrix}\begin{bmatrix}
0 & 1 & 4 & 12 &32 \end{bmatrix}=[97]$
Hence, $(T_2T_1)(2+5x-x^2+3x^4)=T_2(T_1(2+5x-x^2+3x^4))=T_2(5-2x+12x^3)=5-2.2+12.2^3=97$
