Could a real curve symmetric across the line be defined only by polynomial that is not reflection-invariant?

Question

Let $R$ -- be an irreducible plane real algebraic curve (without isolated points).

Suppose that $(x,y)\in R\Leftrightarrow (x,-y)\in R.$

Question: could one find a polynomial $f(x,y)$ with zero set $R$ such that

$$\forall (x,y)\in{\mathbb R^2}\quad f(x,y)=f(x,-y)$$

Answer 1

Let $f$ be any polynomial whose zero set is $R$. Then $F(x,y)=f(x,y)f(x,-y)$ works.