Dimension of certain linear space

I have the linear space $$V = \{ P \in \mathbb{R}_4[x] : xP'''(x) + P''(x) = P'(-1) + P(0) = 0 \}$$ and have to find its dimension. In terms of tools I'm only allowed to use there's the theorem that for any linear $$F : A \to B$$ where $$A$$ and $$B$$ are linear spaces, it's true that $$\dim (\ker (F)) + \dim (\mathrm{Im}(F)) = \dim (A)$$, so my idea is to define some linear $$F$$ such that $$F: \mathbb{R}_4[x] \to \mathbb{R}_4[x]$$, but I have no idea for $$F(P)$$. $$V$$ looks like the kernel of some $$F$$, so it should work, but I don't know how to continue. Could you give me any hints?

You can consider two linear maps: \begin{align} F\colon \mathbb{R}_4[x] & \to \mathbb{R}_4[x], & F(P)&=xP'''(x)+P''(x) \\ G\colon \mathbb{R}_4[x] & \to \mathbb{R}, & G(P)&=P'(-1)+P(0) \end{align} Thus $$V=\ker F\cap \ker G$$.
The matrix of $$F$$ with respect to the basis $$\{1,x,x^2,x^3,x^4\}$$ is $$\begin{bmatrix} 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 12 & 0 \\ 0 & 0 & 0 & 0 & 36 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$ and the matrix of $$G$$ is $$\begin{bmatrix} 1 & 1 & -2 & 3 & -4 \end{bmatrix}$$ Can you finish?
• At the moment there wasn't much talk about bigger matrices than 3x3 on my lecture. However, I think (using theorem I wrote) that $\dim (\ker (F)) = 2$ and $\dim (\ker (G)) = 4$ and now I could use theorem for finite dimensions $\dim (A \cap B) = \dim (A) + \dim (B) - \dim (A+B)$, but what would be the dimension of $A+B$? – chandx Mar 23 at 13:15
• @chandx It's easier to realize that the kernel of $F$ consists of the polynomials having degree at most one; then apply the second condition and you'll have the answer. – egreg Mar 23 at 13:57
• Right, so I see that $\ker (F) = \{ dx + e : d,e \in \mathbb{R} \}$ and second condition tells me that $-4a+3b-2c+d+e = 0$ so in other words $d+e = 0$, thus $e = -d$ and $V = \{ ax - a : a \in \mathbb{R} \}$, so its base is just $\{ x-1 \}$, thus $\dim (V) = 1$, is it right? – chandx Mar 23 at 14:22