Linear polynomial functions with the same Kernel Let $F,G_h\in Lin(\mathbb{R}[x]_{\le3},\mathbb{R}[x]_{\le1})$ be two linear functions such that $F(ax^3+bx^2+cx+d)=(a+c)x+d+2c$, while $G_h(hx^3+\frac{h}2x^2-2x+4)=x+h$ and $G_h(x-1)=1$.
For which values of $h$ $F$ and $G_h$ have the same Kernel?
I found the Kernel of $F$ and it is made up of the span of two vectors. I've not found a way to show that property because I only know how $G_h$ acts on two vectors. Thank you.

We have 
$$\ker F = \text{span}\{x^3-x+2,x^2\} = \text{span}(v_1,v_2)$$
We have to impose some restrictions to $h$ in order to have $\ker G_h = \ker F$.
Called $v_3 = hx^3+\frac{h}{2}x^2-2x+4$ and $v_4=x-1$ we know that 
\begin{gather}
G_h(v_3) = x+h\\
G_h(v_4) = 1
\end{gather}
So $\text{span}(v_3,v_4)\cap \ker G_h = \{0\}$ for all $h$. Since we want $\ker F = \ker G_h$ we have to impose that $\{v_1,v_2,v_3,v_4\}$ is a base of $\mathbb R[x]_{\leq 3}$ (here I used the Grassmann formula: 
$$\dim(\text{span}(v_1,v_2)+\text{span}(v_3,v_4))=\dim(\text{span}(v_1,v_2)) + \dim(\text{span}(v_3,v_4)) -\dim(\text{span}(v_1,v_2)\cap \text{span}(v_3,v_4))
$$
where the last term is $0$ and $v_3,v_4$ are linearly independent for all $h$.)
After the computation, you find that $\{v_1,v_2,v_3,v_4\}$ is a base if and only if $h\neq 2$.
Unfortunatly, we can not do more without other information. Infact for all $h\neq 2$ we can extend arbitrary $G_h$ to a linear application in order to obtain $\ker G_h=\ker F$ or $\ker G_h \neq \ker F$. For, let's take this two extensions:
\begin{equation}
G_h^1(v_1) = 0, \quad G_h^1(v_2)=0, \quad G_h^1(v_3)=x+h, \quad G_h^1(v_4)=1
\end{equation}
and 
\begin{equation}
G_h^2(v_1) = 0, \quad G_h^2(v_2)=1, \quad G_h^2(v_3)=x+h, \quad G_h^2(v_4)=1
\end{equation}
Hence $\ker G_h^1=\ker F$ and $\ker G_h^2 = \text{span}(v_1,v_2-v_4)\neq \ker F$.
