How to calculate roots using root locus method?
You could use Newton's method. Yeah I know, it's not the "root locus method", but hey, it works.



Let \$F(s)=3s^4+10s^3+21s^2+24s+30\$

Clean it up by dividing everything by 3 so our largest s is alone

Let \$f(s)=\dfrac{F(s)}{3}=s^4+\frac{10}{3}s^3+7s^2+8s+10\$

then \$f'(s)=4s^3+10s^2+14s+8\$

Now we have everything we need to start converging onto the values that makes the equation go to 0. 

\$s_{n+1} = s-\dfrac{f(s)}{f'(s)}= s-\dfrac{s^4+\frac{10}{3}s^3+7s^2+8s+10}{4s^3+10s^2+14s+8}\$

Set s to some good initial value and evaluate the equation above until you get a value that stops changing. In other words, you've find a value that makes \$f(s)=0\$. 

If you start with:


\$s = +i\$, then \$s\$ converges to \$-0.059 + 1.525i\$
\$s =-i\$, then \$s\$ converges to \$-0.059 - 1.525i\$


So we have a root pair at \$-0.059\pm1.525i\$



Now, let's find the other two.

So we know that \$f(s)\$ is on the following form because it is a 4th power. 
$$f(s)=(s+0.0589+1.5255i)(s+0.0589-1.5255i)(s+\text{something})(s+\text{something})$$

So let's make a new function called \$f_2(s)\$ which will be equal to:
$$f_2(s)=\dfrac{f(s)}{(s+0.059 + 1.525i)(s+0.059 - 1.525i)}$$

This will cancel our first two roots so Newton's method will converge onto the other two roots. So now we need to find the derivative of the above equation, lovely. Or we just make an approximation and divide by some "large number, say 10. The only difference between dividing by a constant and dividing by the derivative is that it will converge much slower. Let's also reduce the order of \$f'(s)\$ by 2, because the two poles add an order of 2. And remove the terms that has an s to a negative power.

To make it crystal clear what the "optimal" equation would be, it would be:
\$s_{n+1}=s-\dfrac{f_2(s)}{f'_2(s)}\$

But instead the approximation below is used because I'm slightly lazy, and it will converge anyways.

\$s_{n+1} = s-\dfrac{f_2(s)}{10f'(s^{-2})}= s-\dfrac{(s^4+\frac{10}{3}s^3+7s^2+8s+10)}{10(s+0.059 + 1.525i)(s+0.059 - 1.525i)(4s+10)}\$

If you start with:


\$s = +i\$, then \$s\$ converges to \$-1.608 + 1.306i\$
\$s =-i\$, then \$s\$ converges to \$-1.608 - 1.306i\$


So we have a root pair at \$-1.608 \pm1.306i\$



So we can conclude that \$F(s)\$ is equal to:

$$\small F(s)=3(s+0.059+1.525i)(s+0.059-1.525i)(s+1.608+1.306i)(s+1.608-1.306i)$$

Don't forget the factor 3 for \$s^4\$. 



Now, time to verify it. Entering the following code into octave/matlab

h=zpk([(-0.058899+1.525541i) (-0.058899-1.525541i) (-1.608+1.306i) (-1.608-1.306i)],[],3)


returns 

Transfer function 'h' from input 'u1' to output ...

 y1:  3 s^4 + 10 s^3 + 21 s^2 + 24 s + 30.01

Continuous-time model.


Which means that it actually worked, to my surprise.