After Boolean reduction and cancellation over F_2, use exactly the remaining monomial list to form L(X), now adding over the integers. On Boolean x, L(x) counts active monomials and P(x)=0 exactly when L(x) is even. If each monomial has degree at most k and there are s monomials, L has degree at most k and reads at most ks variables. Scope size and degree are different. For P=x+y, L=x+y takes value 2 at (1,1), so this input is a zero of P. Do not cancel the integer lift again modulo two while later evaluating it at rational substituted inputs.