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 $p(x)$).
