Your polynomial $f(x,y)=4x^2+4x-9y^2-1$ cannot be factored (over $\mathbf R$).
Here's a proof. If it could, you could write $f(x,y)=p(x,y)q(x,y)$. Look at the degree of $p$ in $x$: if it's $0$, then all the $x$ is in $q$, meaning $p(x,y)=p(y)$ doesn't depend on $x$. Then $p$ has some root $y_0$ (say in $\mathbf C$), and for all $x$ we would have $f(x,y_0)=0$, which is absurd.
If the degree of $p$ in $x$ is $2$, then all the $x$ is in $p$, and by the same argument, we reach a contradiction, so it has to be $1$. Using another similar argument, we see that the degrees in $y$ of $p$ and $q$ are also both $1$. In other words, you can write $p(x,y)=ax+by+c$ and $q(x,y)=a'x+b'y+c'$, so $f(x,y)$ is a product of two linear polynomials.
Let's put $x=0$ everywhere: we get $-9y^2-1$ for $f(0,y)$, which doesn't have any root in $\mathbf R$, and $(by+c)(b'y+c')$ for $p(0,y)q(0,y)$ which does have roots in $\mathbf R$, because $b,b'\neq0$ (since their product is the coefficient of $y^2$ in $f(x,y)$, and said coefficient is $\neq0$).