Computing Polynomial Dimensions in Linear Transformation 
Let $T:P^3(\Bbb R)→\Bbb R^3$ be the linear transformation defined by
$$T(a+bx+cx^2+dx^3)=(−2c−2b-c-2d, −2a−2b-2d, 4a+4b+3c−2d)$$
Compute the following dimensions:
$\dim(\ker(T))=\cdots $
$\dim(\mathrm{Im}(T))= \cdots$

So, I have this problem to solve but I didn't understand how to do it with polynomials. Can someone help me with it?
 A: Edit: I noticed you say that "you don't know how to do it with polynomials". Polynomials are no different to vectors in the general "vector space" setting. In my answer below, I used the terms "vector" and "polynomial" interchangeably. 
As long as you know what a basis for your space looks like (for polynomials of degree at most three, an obvious choice is $\{1,x,x^2,x^3\}$), you should just think of all the elements of your space as combinations of these basis "vectors". 

Recall that these are related by the dimension theorem. For any operator $T\colon P_3(\mathbb R)\to \mathbb R^3$,
$$\dim(P_3(\mathbb R)) = \dim(\ker T) +\dim(\operatorname{im} T),\tag{$*$}$$
where I'm writing $P_3(\mathbb R)\subseteq\mathbb R[x]$ for polynomials at most degree 3.
We know $\dim(P_3(\mathbb R))$, it's clearly 4, since $\{1,x,x^2,x^3\}$ is a basis and it contains $4$ vectors.
Now I think it's easiest to find the kernel, i.e., the set of vectors (polynomials) such that $Tv=0$. Now write $v=a+bx+cx^2+dx^3$. Then
\begin{align*}
   Tv=0 &\iff T(a+bx+cx^2+dx^3)=0\\
&\iff (−2c−2b-c-2d, −2a−2b-2d, 4a+4b+3c−2d) =0\\
&\iff \text{$c=2d$, $b=-4d$  and $a=3d$}
\end{align*}
Thus a polynomial is in the kernel if and only if it has the form $d+2dx-4dx^2+3dx^3$. This gives us one degrees of freedom, so $\ker T$ has dimension $1$. 
Thus by $(*)$ above, $\dim(\ker T) =1$ and $\dim(\operatorname{im} T) = 3$.
A: So write $T(p(x))=0\implies $ $$(−2c−2d,−2c−2d,b−2a+6c−2d) = 0$$
so $-2c-2d=0\implies d=-c$ and $b-2a+6c-2d=0$ so $b= 2a+8c$. So $$p(x) = ax^3+(2a+8c)x^2+cx-c$$ is in kernel and depends on two parameter, so $\dim(Ker(T))=2$ and by rank-nullity theroem $\dim(Im(T))=2$
