How is this solved ??????????????

s^3 + 6s^2 + 18s + 20 =0
=>(s+2)(s^2 + 4s +10 ) =0

how this 2nd equation is obtained from the 1st equation ????

put s= -2
equate the 1st eq
-8+24-36+20=0
therfore s+2 is a factor of the eq
divide the 1st eq with s+2 you will get s^2+4s+10

check by investigation for which value of s the eqn is equal to 0

in this case it's for s=-2
so, s+2 =0

so (s+2) becomes a factor
other factor will be as^2 + bs +c

so l.h.s of eqn will stand as

(s+2)(as^2 + bs + c)
= as^3 + bs^2 + cs + 2as^2 + 2bs + 2c
= as^3 + (2a+b)s^2 + (2b+c)s + 2c

which is equal to s^3 + 6s^2 + 18s + 20

so we can see a=1 , b=4 , c=10

so the other factor is as^2 + bs + c
= s^2 + 4s + 10

there's another way of doing this. which is a much shorter route. but it has no rule basically you need to imagine this. it's called vanishing method.

the first step will be like below :

s^3 +( )+( )+ 20 = 0

now u need imagination to work out how u will distribute the co-efficients of s^2 and s in the vanished part.