Merging polynomials together

Question

Given the (monic) polynomials $P$ and $Q$, we can split them over an appropriate field into linear factors to write $P(x)=(x-a_1)\dotsb(x-a_m)$ and $Q(x)=(x-b_1)\dotsb(x-b_n)$, and then form the polynomial $$ P(x-b_1)\dotsb P(x-b_n) = Q(x-a_1)\dotsb Q(x-a_m) . $$ Is there any standard notation or name for this new polynomial? Have its properties been studied?

Answer 1

I am going to give an answer to the question asked in the comments

What are alternative ways to define the polynomial in question without factoring the original P and Q

Let $T_x(y)=Q(x-y)$ be understood as a polynomial in $F[x][y]$, where $F$ is a field. I claim that

The resultant $\operatorname{Res}(P(y),T_x(y),y)$ with respect to $y$ is equal to $\prod_{i=1}^mQ(x-a_i)$.

Proof. Let $R$ be an integral domain and $f$ and $g$ monic polynomials in $R[y]$ of degrees $m$ and $n$, and with roots $\lambda_i$ and $\mu_j$ in any algebraically closed field containing $R$, respectively. Their resultant with respect to $y$ is $$\operatorname{Res}(f,g,y)=\prod_{i=1}^m g(\lambda_i)=(-1)^{mn}\prod_{j=1}^n f(\mu_j).$$ Now apply the above for $f(y):=P(y)$ and $g(y):=T_x(y)=Q(x-y)$. The first equality gives $$\operatorname{Res}(P(y),T_x(y),y)=\prod_{i=1}^m T_x(a_i)=\prod_{i=1}^mQ(x-a_i).$$

Corollary. The polynomial $\prod_{j=1}^nP(x-b_j)=\prod_{i=1}^mQ(x-a_i)$ can be computed without factoring $P$ or $Q$.

Proof. Compute the Sylvester determinant for $P(y)$ and $Q(x-y)$ with respect to $y$ (for the relevant definitions see e.g. the wiki articles for Resultant or Sylvester matrix).