Answer
See below
Work Step by Step
From Theorem 6.4.5:
$$[T(p(x))]_C=[T]^C_B[p(x)]_B$$
a) Computing $[T]^C_B$
$T(1)=1.x^2=x^2\\
T(x)=x^2.x=x^3\\
T(x^2)=x^2.x^2=x^4$
then we have $T(1)=x^2=0.1+0.x+1.x+0.x^3+0.x^4\\
T(x)=x^2=0.1+0.x+0.x^2+1.x^3+0.x^4\\
T(x^2)=x^2=0.1+0.x+0.x^2+0.x^3+1.x^4$
Therefore, $[T(1)]_C=\begin{bmatrix}
0 \\ 0 \\ 1 \\ 0 \\ 0
\end{bmatrix}\\
[T(x)]_C=\begin{bmatrix}
0 \\ 0 \\ 0 \\ 1 \\ 0
\end{bmatrix}\\
[T(x^2)]_C=\begin{bmatrix}
0 \\ 0 \\ 0 \\ 0 \\ 1
\end{bmatrix}$
Hence, $[T]_B^C=\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0\\ 1 & 0 & 0 \\ 0& 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$
We can notice that $[p(x)]_B=\begin{bmatrix}
-1 \\
5 \\ 6 \end{bmatrix}$
Consequently, $[T(v)]_C=[T]^C_B[p(x)]_B=\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0\\ 1 & 0 & 0 \\ 0& 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}
-1 \\
5\\6
\end{bmatrix}=\begin{bmatrix}
0 \\
0\\ -1 \\ 5 \\ 6
\end{bmatrix}\\
\rightarrow T(p(x))= -x^2+5x^3-6x^4$
b) From part (a), we obtain: $T(p(x))=T(-1+5x-6x^2)=x^2(-1+5x-6x^2)=-x^2+5x^3-6x^4$
