$Q$ factor of parallel RLC circuit in series with a capacitor and resistor I know that for parallel RLC circuits, the $Q$ factor is given by: 
$$ Q = R \sqrt {\frac{C}{L}} $$
But now suppose it is connected in series to a resistor $R_2$ and capacitor $C_2$. Would the $Q$ factor be changed?

 A: There's a really awesome trick for problems like this.
This is going to be a long post but the method presented makes problems like this really easy.
The idea is to turn the series branch $C_2$, $R_2$ into an effective parallel $R$ and $C$.
See the diagram.
The effective parallel values are denoted $C_{2,p}$ and $R_{2,p}$.
Parallel capacitances just add, so the total capacitance is now $C+C_{2,p}$.
Parallel resistances add in parallel so the total resistance is now $R||R_{2,p} = \left( 1/R + 1/R_{2,p} \right)^{-1}$.
Since we now have a purely parallel circuit, you can stick these values into your formula for $Q$ (which was wrong in the OP, by the way, but I edited it).
Of course, to actually do any of this we have to understand how to solve for $R_{2,p}$ and $C_{2,p}$.
Before we do that I want to simplify some notation.
It is extremely useful to define $Z_{LC} = \sqrt{L/C}$.
This is the "characteristic impedance" of a resonant mode, and it will show up all over the place.
With this definition, the equation for the $Q$ of a parallel $RLC$ resonator is
$$Q = R/Z_{LC}$$
which is really easy to remember:
if $R\rightarrow \infty$ then no current flows through the resistor so there's no energy loss, and as we can see $Q\rightarrow \infty$.

Series/parallel equivalence
Suppose we have a series resistance $R_s$ and reactance $X_s$.
The total series impedance is
$$Z_s = R_s + i X_s .$$
We want to find the equivalent parallel circuit.
The impedance of a parallel resistance $R_p$ and reactance $X_p$ is
$$Z_p = \frac{iR_pX_p}{R_p + iX_p} = \frac{R_pX_p^2 + iR_p^2X_p}{R_p^2 + X_p^2}$$
Now define a new symbol $Q_p \equiv R_p/X_p$.
Using this we can rewrite $Z_p$ as
$$Z_p = \frac{R_p}{1+Q_p^2} + iX_p \frac{Q_p^2}{1+Q_p^2}.$$
Since this is now just a sum of a real number and an imaginary number, it's obvious what the equivalent series values are:
$$R_s = R_p \frac{1}{1+Q_p^2} \qquad X_s = X_p\frac{Q_p^2}{1+Q_p^2} . \qquad (*)$$
Unfortunately we have solved the problem in the wrong direction: we found series values in terms of parallel ones instead of the other way around.
To solve this problem, define $Q_s \equiv X_s/R_s$, and divide the two equations in $(*)$ by one another to find
$$Q_s = Q_p .$$
Now we can easily invert $(*)$ to find
$$R_p = R_s(1 + Q^2) \qquad X_p = \frac{1 + Q^2}{Q^2} X_s$$
where we now write $Q$ instead of $Q_s$ or $Q_p$ because we just showed that they are equal.
We now have the parallel values in terms of the series values.
The best part is that almost always when you have a circuit like the one in the original post, you have $Q \gg 1$, which simplifies the transformation equations considerably to
$$R_p \approx R_s Q^2 \qquad X_p \approx X_s . $$
The take-home message is that the equivalent parallel resistance is transformed to a much larger value, and the equivalent reactance is basically the same as the series value.
Solve the original problem
In the original problem we have
$$
\begin{align}
R_s &= R_2 \\
X_s &= \frac{1}{\omega C_2} \\
Q_e &= \frac{X_s}{R_s} = \frac{1}{\omega C_2 R_2}.
\end{align}
$$
where I've written $Q_e$ to indicate that this is the $Q$ of the "external" circuit.
The equivalent parallel values are
$$
\begin{align}
R_{2,p} &\approx R_2 Q_e^2 \\
X_p &\approx X_s \rightarrow C_{2,p} \approx C_2
\end{align}
$$
We now have a new fully parallel circuit with
$$
\begin{align}
\text{resistance} &= R||R_{2,p} \\
\text{capacitance} &= C + C_{2,p} \approx C \qquad \text{assuming }C \gg C_2 \\
\text{inductance} &= L
\end{align}
$$
The $Q$ of the circuit is
$$
\begin{align}
Q &= \text{resistance} / Z_{LC} \\
&= \left(\frac{1}{R} + \frac{1}{R_2 Q_e^2} \right) ^{-1} / Z_{LC} \\
\frac{1}{Q} &= \frac{Z_{LC}}{R} + \frac{Z_{LC}}{Q_e^2 R_2} \\
&= \frac{1}{Q_i} + \frac{1}{Q_c}
\end{align}$$
where we've defined $Q_i \equiv R / Z_{LC}$ which is the internal $Q$ of the circuit without the external series branch, and $Q_c \equiv Q_e^2 R_2 / Z_{LC}$ is the extra $Q$ induced by the coupling.
In other words, when you add the series branch, the total $Q$ of the resonance winds up being a parallel combination of two components:


*

*$Q_i$: The $Q$ you would have without the coupling to the series branch.

*$Q_c$: The $Q$ you would have if $R$ were absent. This part comes from the coupling to the series branch.
This is, of course, just a result of the fact that $R$ and the effective parallel resistance of the series branch $R_{2,p}$ add in parallel.
We now write down a useful expression for $Q_c$.
First write
$$ Q_e = X_s / R_s = \frac{1}{\omega R_2 C_2} .$$
Since we're talking about properties near resonance, we take $\omega \approx 1/\sqrt{LC}$ giving
$$Q_e = \frac{\sqrt{LC}}{R_2 C_2}.$$
Then for $Q_c$ we get
$$
\begin{align}
Q_c &= \frac{Q_e^2 R_2}{Z_{LC}} \\
&=  \frac{R_2 L C \sqrt{C}}{R_2^2 C_2^2 \sqrt{L}} \\
&= \frac{Z_{LC}}{R_2}\left( \frac{C}{C_2} \right)^2 .
\end{align}
$$
The constitutes a full solution to the problem.
Summary
$$\frac{1}{Q} = \frac{1}{Q_i} + \frac{1}{Q_c}$$
where
$$Q_i = \frac{R}{Z_{LC}}$$
and
$$Q_c = \frac{Z_{LC}}{R_2}\left( \frac{C}{C_2}\right)^2.$$
The approximations made here are that $C_2 \ll C$ and $Q_s \gg 1$.
The approximation $Q_s \gg 1$ is pretty good for $Q_s>3$.
