Suppose $p_0,p_1,\dots,p_m$ are polynomials in $\mathcal P_m(\mathbf F)$ such that $p_j(2) = 0$ for each $j$. Prove that $p_0,p_1,\dots,p_m$ is not linearly indepenent in $\mathcal P_m(\mathbf F)$.

Let $U = \text{span}(p_0,\dots,p_m)$. we know $\dim U \le m+1$

The polynomials $1,(x-2),(x-2)^2,\dots,(x-2)^m$ are independent in $\mathcal P_m(\mathbf F)$ and span $\mathcal P_m(\mathbf F)$ (proof left to the reader) meaning we can write

We see $p(2) = 0$ implies $a_0 = 0$, thus $(x_2),\dots,(x-2)^m$ spans the subspace where $p(2) = 0$, and so by 2.23 $p_0,\dots,p_m$ cannot be linearly independent because it's too long.