Here is another method to solve the Problem.
We are given that #4# is a root of the cubic equation
#p(x)=x^3+2x^2-19x-20=0.#
Let #alpha and beta# be the other #2# roots.
Clearly, #p(x)=(x-4)(x-alpha)(x-beta).#
#:.x^3+2x^2-19x-20=(x-4)(x-alpha)(x-beta).#
Expanding the R.H.S., we get,
#x^3+2x^2-19x-20=x^3-(alpha+beta+4)x^2+(4alpha+alphabeta+4beta)x-4alphabeta.#
Comparing the respective co-effs., we get,
#alpha+beta+4=-2, or, alpha+beta=-6...............<<1>>.#
#4alpha+alphabeta+4beta=-19, i.e.,#
#4(alpha+beta)+alphabeta=-19...................................<<2>>, and,#
#4alphabeta=20, or, alphabeta=5......................................<<3>>.#
Observe that #<<2>># is satisfied by #<<1>> and <<3>>.#
#<<1>> and <<3>># give the desired sum #-6# and the
product #5#.
Enjoy Maths!