Let $p_0,\ldots,p_m \in P_m[\mathbb F]$ be polynomials satisfying $p_i(2) = 0$. Show that the set $\{p_0,\ldots,p_m\}$ is linearly dependent in $P_m[\mathbb F]$.
I have browsed through the website and tried a few things myself. Yet I am not able to find $a_0,\ldots,a_m \in \mathbb F$ s.t.
\begin{equation} a_0p_0(z) + \ldots + a_mp_m(z) = 0(z) = 0 \quad \text{for all} \quad z \in \mathbb F \end{equation}
Am I missing something obvious here?
The 'simple way' (which does not need to talk about the dual space) uses the fact that $\dim_{\mathbb F}\mathbb F_n[x]=n+1$ (I'll use $\mathbb F_n[x]$ for the space of polynomials of degree at most $n$ (together with the zero polynomial) with coefficients in $\mathbb F$.
If $p_0,p_1,\ldots,p_m \in \mathbb F_m[x]$ and $p_i(2)=0 \,\forall i, 0\le i \le m$, then $2$ is a root of each $p_i$, so for each $0\le i \le m$ there are polynomials $q_0,q_1,\ldots,q_m \in \mathbb F_{m-1}[x]$ such that $$p_i(x)=(x-2)q_i(x), \quad 0 \le i \le m$$ (it has to be $\deg q_i \le m-1$ so that $\deg p_i \le m$.)
Now, to prove that $\{p_0,p_1,\ldots,p_m\}$ is a L.D. set, see that $$\alpha_0 p_0(x)+\alpha_1 p_1(x)+\cdots +\alpha_m p_m (x) \equiv 0(x)$$ its equivalent to say that for all $x$ $$\alpha_0 (x-2)q_0(x)+\alpha_1 (x-2)q_1(x)+\cdots +\alpha_m (x-2)q_m (x) =0,$$ and if we have defined the evaluation we can evaluate at $x=3$ so that $$\alpha_0 q_0(x)+\alpha_1 q_1(x)+\cdots +\alpha_m q_m (x) \equiv 0(x),$$ for all $x$.
Now, since all $q_i \in \mathbb F_{m-1}[x]$ (a vector space of dimension $m$) the set of these $m+1$ polynomials is L.D., so there have to be $\alpha_i$ not all of them zero such that the last equation holds. But for those $\alpha_i$ also the first equation holds, which proves $\{p_0,p_1,\cdots,p_m\}$ is L.D.
Also, if you have any knowledge of the concept of dual space, dual base, etc., it's easy to prove that the function $$\epsilon_2 \colon \mathbb F_m[x] \to \mathbb F$$ such that $$\epsilon_2(p)=p(2)$$ (that is, the process of evaluating at $x=2$, or at any other value) is a homeomorphism (a linear transformation), and in particular, a linear form over $\mathbb F_m[x]$. That is, it is an element of the dual space $$\epsilon_2 \in (\mathbb F_m[x])^*.$$ It's clear that $\epsilon_2 \not \equiv 0$, so $T=\langle \epsilon_2 \rangle$ is a subspace of $(\mathbb F_m[x])^*$ with $\dim T=1$.
So if $$S=\{p\in \mathbb F_m[x] \colon p(2)=0\}=\{p\in \mathbb F_m[x] \colon \epsilon_2(p)=0\}=\{p\in \mathbb F_m[x] \colon \phi(p)=0,\, \forall\phi \in T\},$$ by definition this means that $T=Ass(S)$; then $\dim S =\dim \mathbb F_m[x]-\dim T=(m+1)-1=m$.
Then, $m+1$ elements of $S$ form a L.D. set.