Recovering a polynomial from its product with its reciprocal Let $p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in\mathbb{Z}[x]$ be an integer polynomial. We can form the reciprocal polynomial $p^\ast(x)=a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$. The product $p(x)p^\ast(x)$ is a palindromic polynomial.

Can we recover $p(x)$ from $p(x)p^\ast(x)$?

The answer is no, since replacing $p(x)$ with any of $\{p(x),-p(x),p^\ast(x),-p^\ast(x)\}$ does not change the value of $p(x)p^\ast(x)$.

Can we recover $p(x)$ from $p(x)p^\ast(x)$, up to negation and reciprocal?

This can be phrased in fancier language. There is a function $s\colon\mathbb{Z}[x]\to\mathbb{Z}[x]$ given by $p(x)\mapsto p(x)p^\ast(x)$. The Klein four-group $\langle-,\ast\rangle$ acts on $\mathbb{Z}[x]$, and looking at the orbits gives an equivalent relation $\sim$ on $\mathbb{Z}[x]$. Then $s$ is constant on equivalent classes of $\sim$, so we obtain a function $\overline{s}\colon(\mathbb{Z}[x]/\sim)\to\mathbb{Z}[x]$. Is $\overline{s}$ injective?

It is conceivable that the answer is yes, since if you just look at polynomials of degree at most $n$, then this boils down to a system of $n+1$ equations in $n+1$ unknowns.
For example, if you just look at quadratic polynomials then you have to solve the system of equations
\begin{align*}
a_0a_2&=b_0,\\
a_1(a_0+a_2)&=b_1,\\
a_0^2+a_1^2+a_2^2&=b_2.
\end{align*}
Here $b_0,b_1,b_2$ are known (the coefficients of $p(x)p^\ast(x)$), and $a_0,a_1,a_2$ are unknown (the coefficients of $p(x)$).

Not quite.
$$p^*(x)=x^np(1/x)$$
So the roots of $p(x)p^*(x)$ come in pairs $r,1/r$.  Pick one of them for $p$, then the other one goes to $p*$.
If there are no repeated roots, you have $2^n$ choices for $p's$ roots, and I think the sign gives another factor of $2$.
