I have a straightforward question. Let $f$ be a holomorphic cusp form of weight $k$, level $N$, and nebentypus $\chi$ that is new in the sense of Atkin-Lehner theory. Write its Fourier expansion at $\infty$ as $$f(z) = \sum_{n=1}^\infty \lambda_f(n)n^{(k-1)/2}e(nz)$$ and form the $L$-function $L(f,s)$ by the Dirichlet series $$L(f,s)=\sum_{n=1}^\infty \frac{\lambda_f(n)}{n^s}$$ for $\Re(s)>1$. Then $L(f,s)$ can be continued to an entire function on $\mathbf{C}$, and, by the above normalization of Fourier coefficients, obeys a functional equation relating $s$ to $1-s$. Let $\alpha,\beta$ be the Satake parameters associated to $f$; i.e. $$L(s,f) = \prod_p \left(1-\frac{\alpha(p)}{p^s}\right)^{-1}\left(1-\frac{\beta(p)}{p^s}\right)^{-1}$$ and then define the symmetric square $L$-function $L(\operatorname{sym}^2f,s)$ associated to $L(f,s)$ by $$L(\operatorname{sym}^2f,s) := \prod_p (1-\alpha(p)^2p^{-s})^{-1}(1-\alpha(p)\beta(p)p^{-s})^{-1} (1-\beta(p)^2p^{-s})^{-1}.$$ My question is, when exactly is $L(\operatorname{sym}^2f,s)$ entire, and if it is not entire, what poles can it have?

Pages 136–137 of Iwaniec-Kowalski's book seem to answer this question. We know that $L(\operatorname{sym}^2f,s)$ factors like $$L(\operatorname{sym}^2f,s) = L(f\otimes f,s)L(s,\chi)^{-1},$$ where the Rankin-Selberg convolution $L(f\otimes f,s)$ has a simple pole at $s=1$ iff $f$ is self-dual ($f=\overline f$) and is entire otherwise. $L(s,\chi)$ has a simple pole at $s=1$ iff $\chi$ is principal. Therefore, my understanding is that $L(\operatorname{sym}^2f,s)$ is entire unless $f$ is self-dual with non-principal character. This can happen if $f$ is a `CM form' arising from a Hecke grössencharacter; see https://mathoverflow.net/a/164126/37110 for details. In the special case when $f$ has real Fourier coefficients and $\chi$ is not principal, $L(\operatorname{sym}^2f,s)$ has a simple pole at $s=1$.

Is my understanding correct? Is it complete? Thanks in advance.