Let $T: M_{2x2} \rightarrow P_3$ be the linear transformation
$$T( \begin{bmatrix} a & b \\ c & d \end{bmatrix})=(a+c)x^3+(a+b+c)x^2+(d-a-c)x+(a+b+c+d)$$
a) Find bases for the kernel and range of $T$ and compute their dimensions. For the kernel, do I make all of the coeffients=0? Like $(a+c)=0, (a+b+c)=0$... and solve? If so, I just get that $a=-c, b=0,$ $c:$ free,$d=0,$ so would the basis for kernel be the matrix \begin{bmatrix} -1 & 0 \\ 1 & 0 \end{bmatrix} with dimension 1? For the range, I reduced the same matrix that I did for the kernel, found the pivot columns (columns 1,2, and 4) and wrote the basis as the set of those original columns: {(x^3+x^2-x+1),(x^2+1),(x+1)}. Is this right? I'm just skeptical since I can't find a way to get the polynomial "x" from that set.
b) Determine whether or not $T$ is one-to-one and/or onto. If I found the right kernel then this would not be one-to-one since the kernel isn't 0, but for the range, how do I know if it is onto? Since there wasn't a pivot in every row of the matrix I reduced in part (a), does that mean that it is not onto?
For the kernel, you are right and just have to explicitly write down the set.
For the range, a hint is to write the polynomial as $$(a+c)x^3+(a+b+c)x^2+(d−a−c)x+(a+b+c+d)$$ $$=a(x^3+x^2-x+1)+b(x^2+1)+c(x^3+x^2-x+1)+d(x+1)$$ and try to see what the image could be.
In summary, by your calculation, we can have $$\text{Ker}~T=\{\begin{bmatrix} -t & 0 \\ t & 0 \end{bmatrix}|t\in\mathbb{R}\}$$ which is a one-dimensional subspace.
Besides, the image of $T$ is spanned by the basis $\{x^3+x^2-x+1,x^2+1,x+1\},$ so the image is of dimension $3$ while the dimension of $P_3$ is 4, and hence $T$ is not onto.