in an equation of type #x^n+a_1x^(n-1)+a_2x^(n-2)+.........+a_n#, the roots of the equation are factors of #a_n#.
Here, we have the equation as #x^3-8x-3=0# and hence roots are factors of #3# i.e. #+_1# and #+-3#.
It is apparent that #x=3# is a root as it satisfies the equation
#3^3-8xx3-3=27-24-3=0# and hence
#x-3# is a factor of #x^3-8x-3# and dividing latter by former
#x^2(x-3)+3x(x-3)+1(x-3)=(x-3)(x^2+3x+1)#
As such we have #(x-3)(x^2+3x+1)=0#
And as #x^2+3x+1# cannot be factorized as rational factors, using quadratic formula roots are
#(-3+-sqrt(3^2-4xx1xx1))/2=(-3+-sqrt5)/2#
and hence three roots of #x^3-8x-3=0# are #3#, #(-3+sqrt5)/2# and #(-3-sqrt5)/2#
