A real polynomial vanishing at $a$ is divisible by $x-a$

For each $k\ge 1$, the power-difference factorization gives $x^k-a^k=(x-a){\sum }_{j<k}{a}^j{x}^{k-1-j}$.

Since $p(a)=0$, write $p(x)={\sum }_{k<n}{a}_k(x^k-a^k)$.

If the displayed list has length $n\le 1$, the root condition makes its constant evaluation zero, so choose the zero polynomial $q$. For $n\ge 2$, define $b_d={\sum }_{k=d+1}^{n-1}{a}_k{a}^{k-1-d}$ for $0\le d\le n-2$. Reindexing the finite sums in steps 1.1–1.2 gives $p(x)=(x-a){\sum }_{d=0}^{n-2}{b}_d{x}^d$. If every $b_d$ vanishes choose the zero formal polynomial; otherwise truncate the list $(b_0,\dots ,b_{n-2})$ after its last nonzero entry. This is a formal polynomial $q$ with the required evaluation. In particular, when $p$ has degree $N\ge 1$ and its actual list is $(a_0,\dots ,a_N)$, take $n=N+1$: then $b_{N-1}=a_N{a}^0=a_N\mathrm{\ne }0$, and no exponent above $N-1$ occurs. Thus this construction gives $q$ of degree exactly $N-1$ and leading coefficient $a_N$.