Bode plot of transfer function?
You need to calculate the roots of numerator and denominator.

For the numerator you have \$(s^2+s+1)=0 \Leftrightarrow s_{n_{1,2}}=-\frac{1}{2} \pm j\sqrt{\frac{3}{4}}\$

For the denominator you have \$(s+1) =0 \Leftrightarrow s_{d_{1}}=-1\$ as well as 
$$(2 s^2+1) =0 \\
\Leftrightarrow s^2=- {\frac{1}{2}}\\
\Leftrightarrow s_{d_{2,3}}=\pm j\sqrt{\frac{1}{2}}
$$
because a product is zero if one or more of its factors are zero.

Therefore your transfer function is
$$
W(s)=\frac{(s-s_{n_{1}}) \cdot (s-s_{n_{2}})}{(s-s_{d_{1}})\cdot(s-s_{d_{2}})(s-s_{d_{3}})}
$$

Here is a picture of what the bode plot should look like (sorry for just using Matlab instead of drawing by hand):


Please note that the transfer function drops with 20dB/decade after the resonance.

As for the resonance: @Mario has explained very well that you see a resonance peak because of the complex conjugated pole pair on the imaginary axis. Because it is in the denominator, the peak points upwards. As there is no dampening (the peak is on the imaginary axis), you get an arbitrary high value.