G(S)H(S)=$\frac{120}{(S+2)(S+10)}$

Arrange the above equation in time control form,

G(S)H(S)=$\frac{120}{(2(1+0.5S)}$ 10(1+0.1S))

G(S)H(S)=$\frac{6}{(1+(\frac{S}{2}) (1+\frac{S}{10})}$

Comparing with standard form of time constant,

K=6

Mo pole at origin.

Simple pole at $\frac{1}{(1+(S+2)}$, $w_c1$=2 rad/sec

Simple pole at $\frac{1}{(1+(S+10)}$, $w_c2$=10 rad/sec

Magnitude plot:

For k=6, 20log6=15.563 dB

Phase angle plot:

G(jw)H(jw)=$\frac{6/(1+j(w/2)}{(1+j(w/10)}$

=$\frac{\lt6+j0)}{\lt1+j(w/2)(\lt1+j(w/10)}$

The table for phase angle will be:

From Bode plot as shown in the figure:

GM=∞; PM=58

Hence the system is stable.