My question is related to this: Can we remove a root of a polynomial via multiplication?
Let all operations and values defined over a finite field $\mathbb{F}_p$, where $p$ is a large prime number. Also, all polynomials are defined over $\mathbb{F}_p[x]$. We assume all values and polynomials are non-zero.
Let $F(x)=G(x)\cdot (x-\beta)$ where $\beta$ is an arbitrary element of the field.
We want to remove a root of $F(x)$ via multiplication. We set $D(x)=\frac{S(x)}{x-\beta}$.
Question: Is still $\beta$ a root of $R(x)=D(x)\cdot F(x)$?
Note that $(x-\beta)$ does not divide $S(x)$.
Edit:
Can I prove the following corollary as follows: Let $\tau(x)$ and $\gamma(x)$ be two arbitrary non-constant polynomials of degree at most $d$ such that $\tau(x),\gamma(x)\in \mathbb{F}_p[x]$. Then, the product of the two polynomials does not cancel out the polynomials roots, so the roots of $\alpha(x)=\tau(x)\cdot\gamma(x)$ include all roots of both $\tau(x)$ and $\gamma(x)$.
proof: Let $\gamma(x)=\prod\limits^{b}_{i=1 }(x-c_{i})\prod \limits^{d-b}_{j=1}(x-e_{j})$, where $1\leq b\leq d$. The only way to remove $b$ roots, e.g. $c_{i}$, from $\gamma(x)$ via polynomial multiplication is either to set $\tau(x)$ to $\frac{\sigma(x)}{\prod\limits^{b}_{i=1}(x-c_{i})}$ for any non-zero $\sigma(x)$, not divisible by the denominator. But, $\frac{\sigma(x)}{\prod\limits^{b}_{i=1}(x-c_{i})}$ is undefined in $\mathbb{F}_p[x]$. Or, to set $\tau(x)$ to the multiplicative inverse of $\prod\limits^{b}_{i=1}(x-c_{i})$. Nonetheless, there is no non scallar polynomials in $\mathbb{F}_p[x]$ that has a multiplicative inverse. Thus, the product of the two polynomials preserves all roots of both.