Suppose that $p_0, p_1, ..., p_m$ are polynomials in $P_m(F)$ such that $p_i(2)=0$ for each $i$. Prove that the set of vectors $p_0, ..., p_m$ is linearly dependent.
If each $p_i(t)$ evaluates to $0$ at $t = 2$, why can you conclude that the $p_i$ must therefore be linearly dependent?:
For intuition: Try to construct linearly independent polynomials, say for $P_2(F)$, such that for each, $p_i(2) = 0$; and see why these polynomials cannot, in fact, be linearly independent.
Suppose there are $m+1$ linearly independent polynomials such that $p_i(2) = 0 \forall p_i$. Then $p_0 = 0$. Hence, e.g., 1 would necessarily be linearly independent of the $m+1$ $p_i$, also in $P_m(F)$. That would give a basis of $P_m(F)$ that has $m+2$ linear independent vectors. Impossible!