I am using polynomials to represent bit vectors, such that the value of the d-th coefficient is $1$ if the d-th bit in the vector is $1$, and $0$ otherwise. For example, the vector $A = [1, 1, 0, 1, 0]$ is represented by the polynomial $P_A(x) = 1 + 1x + 0x^2 + 1x^3 + 0x^4 = 1 + x + x^3$.
I understand that I can find the number of bits $1$ in the bitwise AND of two vectors, $|A \wedge B|$, by multiplying $P_A(x)$ and the "inverted" $P_B(x)$ and checking the (d-1)-th coefficient in the resulting polynomial, as explained in Scalar product of vectors over polynomial rings.
Now I want a similar way to find the number of bits $1$ in the bitwise OR: $|A \vee B|$. Considering $|A \vee B| = |A| + |B| - |A \wedge B|$, so far I thought about of $|A|$ by multiplying $P_A(x) \cdot P_1(x)$, where $P_1(x)$ is the polynomial having the value $1$ for all coefficients. Then, I would do the same to find $|B|$.
This would require three polynomial multiplications and two additions: $P_A(x) \cdot P_1(x) + P_B(x) \cdot P_1(x) + P_A(x) \cdot P_B(x)$. Since the first two have $P_1(x)$ in common, it is possible to simplify it a bit: $(P_A(x) + P_B(x)) \cdot P_1(x) + P_A(x) \cdot P_B(x)$, using a total of two polynomial multiplications.
My question is: Is there a way to do this using fewer polynomial multiplications?