Well, when we have a transfer function that looks like:
$$\mathcal{T}:=\frac{\text{Y}\left(\text{s}\right)}{\text{X}\left(\text{s}\right)}=\frac{\text{G}\left(\text{s}\right)}{1+\text{G}\left(\text{s}\right)\text{H}\left(\text{s}\right)}\tag1$$
Where $\text{H}\left(\text{s}\right):=1$ because of the unity feedback and $\text{G}\left(\text{s}\right):=\text{K}\cdot\frac{1}{\text{s}-\alpha}$
We get a complete transfer function that looks like:
$$\mathcal{T}=\frac{\text{Y}\left(\text{s}\right)}{\text{X}\left(\text{s}\right)}=\frac{\text{K}\cdot\frac{1}{\text{s}-\alpha}}{1+\text{K}\cdot\frac{1}{\text{s}-\alpha}}\tag2$$
Now, for the poles and zeros:
- Poles:
$$1+\text{K}\cdot\frac{1}{\text{s}-\alpha}=0\space\Longleftrightarrow\space\text{s}=\alpha-\text{K}\tag3$$
- Zeros:
$$\text{K}\cdot\frac{1}{\text{s}-\alpha}=0\space\Longleftrightarrow\space\text{K}=0\tag4$$
For the settling-time we have that:
$$t\space_{2\text{%}}=-\frac{\ln\left(50\right)}{\lambda}\space\Longleftrightarrow\space\lambda=-\frac{\ln\left(50\right)}{t\space_{2\text{%}}}\tag5$$
We need to use:
$$\text{s}=\lambda+\omega\text{j}\tag6$$
Where $\lambda\space\wedge\space\omega\in\mathbb{R}$ and $\text{j}^2=-1$
Now, we need to solve for $\text{K}$ in the pole equation (assuming $\text{K}\ne0$):
$$1+\text{K}\cdot\frac{1}{\left(-\frac{\ln\left(50\right)}{t\space_{2\text{%}}}+\omega\text{j}\right)-\alpha}=0+0\text{j}\space\Longleftrightarrow\space\text{K}=\alpha+\frac{\ln\left(50\right)}{t\space_{2\text{%}}}\space\space\space\wedge\space\space\space\omega=0\tag7$$
Conclusion, we get a transfer function that looks like:
$$\color{red}{\mathcal{T}=\frac{\text{Y}\left(\text{s}\right)}{\text{X}\left(\text{s}\right)}=\frac{\alpha\cdot t\space_{2\text{%}}+\ln\left(50\right)}{\text{s}\cdot t\space_{2\text{%}}+\ln\left(50\right)}}\tag8$$
Notice that
$$
\require{cancel}
\frac{10 s^3 + 70 s^2 + 200 s + 500}{s^5 + 9 s^4 + 44 s^3 + 210 s^2 + 400 s + 1000} = \frac{(10 s + 50)\cancel{(s^2 + 2s + 10)}}{(s^3 + 7 s^2 + 20 s + 100)\cancel{(s^2 + 2s + 10)}}
$$
What you see happens because of numeric inaccuracy so the pole/zero cancellation is not done.
Best Answer
The closed-loop transfer function is, \begin{align} y &= G(x-z) \\ &= G(x-Hy) \\ (1+GH)y &= Gx \\ y &= \frac{G}{1+GH}x \end{align} Let $n_G$, $n_H$ be the numerator of $G$ and $H$, and $d_G$, $d_H$ be the denominator of $G$ and $H$.
\begin{align} y &= \frac{G}{1+GH}x \\ &= \frac{\frac{n_G}{d_G}}{1+\frac{n_G}{d_G}\frac{n_H}{d_H}}x \\ &= \frac{n_Gd_H}{n_Gn_H + d_Gd_H}x \\ &= \frac{K(s-1)(s-2)}{K(s-1)(s-2) + (s+1)(s+2)}x \\ &= \frac{K(s-1)(s-2)}{K(s^2-3s+2) + (s^2+3s+2)}x \\ &= \frac{K(s-1)(s-2)}{(1+K)\left[s^2+\frac{3(1-K)}{1+K}s+\frac{2+K}{1+K}\right]}x \end{align}
$\omega_n^2 = \frac{2+K}{1+K}$ and $2\zeta\omega_n = \frac{3(1-K)}{1+K}$. Therefore, \begin{align} \zeta &= \frac{3(1-K)}{2\omega_n} = \frac{3(1-K)}{2\sqrt{2+K}\sqrt{1+K}} \end{align}
If you want $\zeta = \frac{1}{\sqrt{2}}$ then $K= \frac{12-\sqrt{109}}{7} $