Calculating vertex factor for scalar field theory
I am practising basic QFT and am having some trouble with calculating the vertex factor of an interacting theory involving two real scalar fields, $\phi_{1}$ and $\phi_{2}$.
If I create a generic Lagrangian with interaction terms:
$$\mathcal{H}{\text{int}}=g\phi{1}^{2}\phi_{2} + q\phi_{1}^{2}\phi_{2}^{2}$$
I want to calculate the vertex factors associated with these interactions. So to calculate the $\phi_{1}^{2}\phi_{2}$ vertex, I calculate the three-point Green's function:
$$G^{(3)}(x,y,z)=\langle\Omega|T\phi_{1}(x)\phi_{1}(y)\phi_{2}(z)|\Omega\rangle = \frac{\langle0|T\phi_{1}(x)\phi_{1}(y)\phi_{2}(z)\exp\left(-i\int_{-\infty}^{\infty}\mathcal{H}{\text{int}}\:\mathrm{d}^{4}s\right)|0\rangle}{\langle0|T\exp\left(-i\int{-\infty}^{\infty}\mathcal{H}_{\text{int}}\:\mathrm{d}^{4}s\right)|0\rangle}$$
We can expand the exponential involving $\mathcal{H}_{\text{int}}$ to first order in $g$ and $q$:
$$\exp\left(-i\int_{-\infty}^{\infty}\mathcal{H}{\text{int}}\:\mathrm{d}^{4}s\right)=1-ig\int{-\infty}^{\infty}\phi_{1}^{2}(s)\phi_{2}(s)\:\mathrm{d}^{4}s-iq\int_{-\infty}^{\infty}\phi_{1}^{2}(s)\phi^{2}_{2}(s)\:\mathrm{d}^{4}s$$
Clearly by <PERSON>'s theorem, any term without an even number of $\phi_{1}$ or $\phi_{2}$ terms will disappear, so the denominator will become:
$$\langle0|T\exp\left(-i\int_{-\infty}^{\infty}\mathcal{H}{\text{int}}\:\mathrm{d}^{4}s\right)|0\rangle = 1-iq\int{-\infty}^{\infty}\langle0|\phi_{1}^{2}(s)\phi_{2}^{2}(s)|0\rangle\:\mathrm{d}^{4}s = 1 - iq\int_{-\infty}^{\infty}\:\mathrm{d}^{4}s\Delta_{F}^{(1)}(0)\Delta_{F}^{(2)}(0)$$
Where $\Delta_{F}^{(1)}(z)$ is the Feynman propagator for $\phi_{i}$.
My understanding is (from this other question I asked), that terms containing a tadpole diverenge ($\Delta^{(i)}{F}(0)$) can be set to zero using the requirement that $\langle \phi{i}\rangle = 0$, so the denominator evaluates to $1$. 281