Linear Transformation in $\mathbb R_2[X]$ I'm learning linear algebra, specifically linear transformations, and need help with the following exercise:

Consider the linear transformation $f:\mathbb R_2[X] \rightarrow \mathbb R_2[X]$ satisfying the relations 
\begin{align}
f(1+X+X^2)&=1+X+2X^2\\
f(X+X^2)&=1+3X+6X^2\\
f(X^2)&=X+4X^2
\end{align}
and let $Q(X) = 4+10X+\mu X^2, \mu \in \mathbb R$. 
$(\alpha)$ Determine $f$.
$(\beta)$ Determine the value of $\mu$ for which $Q(X)$ belongs to $\operatorname{Im}f$. Also, find all the $P(X) \in \mathbb R_2[X]$ such that $f(P(x)) = Q(x)$. 

Since I'm having difficulties for $(\beta)$, I'm going to share my thoughts for $(\alpha)$. 
$(\alpha)$ We consider the standard basis $\mathscr{B}=\{1,X,X^2\}$ of $\mathbb R_2[X]$. Given that $f$ is a linear transformation, it can be represented by a matrix whose columns are the images of the standard basis vectors. The third column of the requested matrix is found immediately. It is given by 
$$f(X^2)=X+4X^2.$$ 
By linearity, we determine the first two columns. One has
\begin{align}
f(1) &= f\left((1+X+X^2) - (X+X^2)\right) = f(1+X+X^2) - f(X+X^2)\\
&= (1+X+2X^2) - (1+3X+6X^2) = -2X - 4X^2
\end{align}
and 
\begin{align}
f(X) &= f\left((X+X^2) - (X^2)\right) = f(X+X^2) - f(X^2)\\
&= (1+3X+6X^2) - (X+4X^2) = 1+2X+2X^2.
\end{align}
Hence, the matrix of $f$ relative to the basis $\mathscr{B}$ is
$$A = \begin{pmatrix}0 & 1 & 0\\-2 & 2 & 1\\-4 & 2 &4\end{pmatrix}$$
Is my work correct for $(\alpha)$? Since we are asked to determine $f$, I'm not sure if my answer is complete. I only determined the matrix representation of $f$. Can I explicitly express $f$ from here? Also, any help for $(\beta)$ would be appreciated. I can state the obvious by saying that $\operatorname{Im}f = \operatorname{Col}A$ but this does not help a lot. 

I would say your solution for $\alpha$ is one of the equivalent ways of expressing $f$. Another would be to say that $$f(ax^2+bx+c) = a(4x^2+x) + b(2x^2+2x+1) + c(-4x^2-2x)$$
and then simplify that expression.

For $\beta$, you already have the matrix set up, so the question is for which value of $\mu$ does the equation $$\begin{pmatrix}0 & 1 & 0\\-2 & 2 & 1\\-4 & 2 &4\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix} = \begin{pmatrix}4\\10\\\mu\end{pmatrix}$$
have a solution, and since the matrix has full rank, the answer should be simple.
