x^5+2x^4-3x^3-8x^2-4x = 0
The first factor we can take out is x.
x(x^4 + 2x^3 - 3x^2 - 8x - 4) = 0
By inspection x = -1 is a root, hence (x+1) is a factor. Taking out this factor,
x(x+1)(x^3 + x^2 - 4x - 4) = 0
By inspection x = -1 is again a root, hence (x+1) is again a factor, Taking out this factor,
x(x+1)(x+1)(x^2 - 4) = 0
The end term is a difference of two squares and factorises as,
x(x+1)^2(x-2)(x+2) = 0
The roots of the equation are: x = 0, x = -1 (twice), x = 2, x = -2