Consider a $k$-homogeneous polynomial $p:\Bbb R^n \to \Bbb R$ defined by $$ p(\xi)=\sum_{|\alpha|=k}c_\alpha \xi^\alpha, $$ where $\alpha=(\alpha_1,\dots,\alpha_n)$ is a multiindex. Here $\xi^\alpha = \xi_1^{\alpha_1}\cdots \xi_n^{\alpha_n}$ and $|\alpha| = \alpha_1+\dots+\alpha_n$.
Suppose that $\eta\in S^{n-1}$ is a root of $p$, i.e. $p(\eta)=0$. For a rotation matrix $R\in \text{SO}(n)$, we may defined the rotated polynomial $\tilde p$ by $$ \tilde p(\xi) = p(R\xi). $$ The matrix $R$ is chosen so that $\eta = Re_1$, where $e_1=(1,0,0,\dots,0)$, hence we have $$ \tilde p(e_1) = 0. $$
Expanding out $\tilde p$ into the summation form yields $$ \tilde p(\xi) = \sum_{|\alpha|=k}\tilde c_\alpha \xi^\alpha. $$ Is it true that $\tilde c_\alpha =0$ whenever $\alpha_1\ne 0$? (i.e. there is no term containing $\xi_1$)
As an easy example, consider $p:\Bbb R^2\to \Bbb R$ where $p(\xi)=\xi_1-\xi_2$. We have $\eta = \frac 1{\sqrt{2}}(1,1)$ is a root of $p$ and we can take $R =\frac 1{\sqrt 2}\pmatrix{1 &-1\\ 1 &1}$. This gives $\tilde p(\xi) = -\sqrt 2 \xi_2$.
PS. Forgive me if my tags seem a bit random, I am not quite sure myself which scope should this question belong to. I had this question when trying to transform some symbol of a partial differential operator but I have very limited knowledge in geometry.
Let's assume we have a homogeneous polynomial $$\tilde{p}(\xi) = \sum_{|\alpha| = k} \tilde{c}_{\alpha} \xi^{\alpha} $$ such that $\tilde{p}(e_1) = 0$. Note that $$ e_1^{\alpha} = (e_1)_1^{\alpha_1} \dots (e_1)_n^{\alpha_n} = \begin{cases} 0 & \alpha_1 \neq k, \\ 1 & \alpha = (k,0,\dots,0)\end{cases} $$ where $(e_1)_i = \delta_{i,1}$ is the $i$-th coordinate of $e_1$. Hence, $$ \tilde{p}(e_1) = \tilde{c}_{(k,0,\dots,0)} = 0. $$ In general, that's all you can deduce. For example, if $n = 2$, the second degree homogeneous polynomial $$ \tilde{p}(\xi) = \xi_1 \xi_2 $$ satisfies $\tilde{p}(e_1) = 0$ but it has a term involving $\xi_1$.