Problem
Suppose $p_0, p_1, \dots, p_m$ are polynomials in $\mathcal{P}_m(\mathbb{F})$ such that $p_j(2) = 0$ for each $j$. Prove that $p_0, p_1, \dots, p_m$ is not linearly independent in $\mathcal{P}_m(\mathbb{F})$
Solution
We know that $z-2 | p_i$ for all $i$, therefore we now denote $q_i$ as the polynomial $p_i/(z-2)$. Therefore: $$0 = a_0p_0 + a_1p_1 + \dots + a_mp_m$$ Is only true in general if and only if: $$0 = a_0q_0 + a_1q_1 + \dots + a_mq_m$$ But $q_i$ has degree at most $m-1$ and the linearly independent list $1,z,\dots,z^{m-1}$ spans $\mathcal{P}_{m-1}(\mathbb{F})$, therefore by theorem 2.23 $q_0, q_1, \dots, q_m$ is linearly dependent implying that there exist $a_0, a_1, \dots, a_m$, not all zero, such that $0 = a_0q_0 + a_1q_1 + \dots + a_mq_m$ and by extension $0 = a_0p_0 + a_1p_1 + \dots + a_mp_m$.