Answer
See below
Work Step by Step
Obtain $T(1)=(1,1)\\ T(x)=(0,1) \\T(x^2)=(0,1) \\ T(x^3)=(0,1)$
then we have:
$T(1)=(1,1)=1.(1,0)+1.(0,1)\\
T(x)=(0,1)=0.(1,0)+1.(0,1)\\
T(x^2)=(0,1)=0.(1,0)+1.(0,1)\\
T(x^3)=(0,1)=0.(1,0)+1.(0,1)$
as $[T(1)]_C=\begin{bmatrix}
1\\
1
\end{bmatrix}\\
[T(x)]_C=\begin{bmatrix}
0 \\ 1
\end{bmatrix} \\
[T(x^2)]_C=\begin{bmatrix}
0 \\ 1
\end{bmatrix} \\ [T(x^3)]_C=\begin{bmatrix}
0 \\ 1
\end{bmatrix}$
Hence, $[T]_B^C=\begin{bmatrix}
1 & 0 & 0 & 0 \\
1 & 1 & 1 & 1\end{bmatrix}$
We can notice that $[v]_B=\begin{bmatrix}
2 & 0 & -1 & 2\end{bmatrix}$
From Theorem 6.4.5,
$$[T(v)]_C=[T]^C_B[v]_B\\
\rightarrow [T(v)]_C=\begin{bmatrix}
1 & 0 & 0 & 0 \\
1 & 1 & 1 & 1\end{bmatrix}\begin{bmatrix}
2 & 0 & -1 & 2\end{bmatrix}=\begin{bmatrix}
2 & 3\end{bmatrix}\\
\rightarrow T(2-x^2+2x^3=(2,3)$$
