This one is pretty straightforward. We set K1 = 4 and want to bring out K2 as a common factor so we can plot the root locus for the system. The correct answer is:
$$\small G_{new}(s)=4K_2\frac{(s+6)}{(s-2)(s+6)+20}$$
I can't think of any algebraic trick to get this result.
To simplify calculations start with substituting \$a=s+6\$ and \$b=s-2\$.
The closed-loop transfer function is \$\frac{g}{1+g}\$, where \$g=k1(\frac{5}{a}+k2)\frac{1}{b}=\frac{k1 (5+a\ k2)}{a\ b}\$.
Thus the closed-loop transfer function \$\frac{g}{1+g}\$ becomes \$\frac{k1 (5+a\ k2)}{a\ b+k1 (5+a\ k2)}\$.
The closed-loop poles are obtained from the denominator of the closed-loop transfer function. That is,
$$a\ b+k1 (5+a\ k2)=0$$
$$a\ b+k1 \ 5+k1\ a\ k2=0$$
$$1+\frac{k1\ a\ k2}{a\ b+k1 \ 5}=0$$
Now substitute \$k1=4\$, \$a=s+6\$, and \$b=s-2\$, to get
$$1+4\ k2 \frac{s+6}{(s+6)(s-2)+20}=0$$
