Finding the action of T on a general polynomial given a basis I am given a question as follows:

Suppose $T: P_{2} \rightarrow M_{2,2}$ is a linear transformation whose action on a basis for $P_{2}$ is 
  $$T(2x^2+2x+2)=\begin{bmatrix} 2&4 \\ 2&8 \end{bmatrix}  \\T(4x^2+2x+2)=\begin{bmatrix}4&6 \\ -2&10 \end{bmatrix} \\T(1)=\begin{bmatrix} 1&1 \\ 2&1 \end{bmatrix}.$$
Find the action of $T$ on a general polynomial with $a,b,c$ as constants. 

Assuming that $T$ is a linear transformation, then for any vectors $\vec{v_{1}}, ..., \vec{v_{n}}$ and any scalars $a_{1}, ..., a_{n}$, $T(a_{1}\vec{v_{1}}+...+a_{n}\vec{v_{n}})=a_{1}T(\vec{v_{1}})+...+a_{n}T(\vec{v_{n}})$.
The answer is supposed  to be in matrix form.
What I have tried doing so far is taking the basis polynomials and putting them into a matrix, which I then multiplied by a constant vector consisting of $a,b,c$:
$$\begin{bmatrix} 2&4&0 \\ 2&2&0 \\ 2&2&1\end{bmatrix} \begin{bmatrix} a \\ b \\c \end{bmatrix}= \begin{bmatrix} 2a+4b \\ 2a+2b \\ 2a+2b+c \end{bmatrix}$$
Given what I have done so far, I am not sure if this is the right approach, and I am not sure where to proceed from here. Any help would be appreciated!
 A: First determine the matrix representation relative to the basis, $T(1), T(x),T(x^2)$.$$T(x^2)=\frac{T(4x^2+2x+2)-T(2x^2+2x+2)}{2}\\=(\begin{bmatrix}4&6 \\ -2&10 \end{bmatrix}-\begin{bmatrix} 2&4 \\ 2&8 \end{bmatrix})/2 =\begin{bmatrix}1&1 \\ -2&1 \end{bmatrix}$$
$$T(x)=\frac{T(2x^2+2x+2)-2T(x^2)-2T(1)}{2}=(\begin{bmatrix} 2&4 \\ 2&8 \end{bmatrix}-2\begin{bmatrix}1&1 \\ -2&1 \end{bmatrix}-2\begin{bmatrix} 1&1 \\ 2&1 \end{bmatrix})/2=\begin{bmatrix} -1&0 \\ 1&2 \end{bmatrix}$$
$$T(1)=\begin{bmatrix} 1&1 \\ 2&1 \end{bmatrix}$$Therefore,$$T(ax^2+bx+c)=aT(x^2)+bT(x)+cT(1)=\begin{bmatrix} a-b+c&a+c \\ -2a+b+2c&a+2b+c \end{bmatrix}$$
A: 
Hint:
Let $p_1 = 2x^2 + 2x + 2$, $p_2 = 4x^2 + 2x + 2$, $p_3 = 1$. Then
  $$
-\frac{1}{2}p_1 + \frac{1}{2}p_2 = x^2, \\
p_1 -\frac{1}{2}p_2 - p_3 = x, \\
p_3 = 1.
$$

Thus 
$$
ax^2 + bx + c = \left(-\frac{a}{2} + b\right)p_1 + \frac{a-b}{2}p_2 + (c - b)p_3
$$
and as $T$ is a linear operator we get
$$
T(ax^2 + bx + c) = \left(-\frac{a}{2} + b\right)T(p_1) + \frac{a-b}{2}T(p_2) + (c - b)T(p_3) = \\
\left(-\frac{a}{2} + b\right)\begin{bmatrix} 2&4 \\ 2&8 \end{bmatrix}  + \frac{a-b}{2}\begin{bmatrix}4&6 \\ -2&10 \end{bmatrix} + (c - b)\begin{bmatrix} 1&1 \\ 2&1 \end{bmatrix} = \\
= \begin{bmatrix} a-b+c&a+c \\ -2a+b+2c&a+2b+c \end{bmatrix}.
$$
