Use trial and error to find 2 integer zeroes of F(x) then use sum and product of zeroes to find any other zeroes. Note any multiple zeroes.
F(x)=x^4-6x^2-8x-3
I don't know how to go about this
any zeroes of polynomial must be factors of the component of the polynomial that has no x in it ie the 3. Thus sub in 1, 3, etc and if not work try fractions like 1/2, 1/3 etc until u get 2, and use sum and product of roots with alpha, beta and the 2 u found to make alpha+beta+x+y=0 (minus b/a, and b is x^3 which is 0) and alpha.beta= -6/4 (c/a), so by quadratic x^2-Sx+p, the other two zeroes can be found by saying x^2-(x+y)-(6/4)
