To find zeros of #f(x)=2x^3+x^2-8x-4#, first identify factors #-4#, for which #f(x)=0#. If its zero for #x=a#, #(x-a)# will be a factor of #f(x)#.

Factors of #-4# are #{1,-1,2,-2,4,-4}# and it is apparent that for #x=2#, #f(2)=2(2)^3+2^2-8*2-4=16+4-16-4=0#, hence #(x-2)# is a factor of #2x^3+x^2-8x-4#. Dividing latter by former

#2x^3+x^2-8x-4=(x-2)(2x^2+5x+2)#

Now #(2x^2+5x+2)# can be easily factorized by splitting #5x# as #4x+x# - (In a quadratic polynomial #ax^2+bx+c#, one has to identify factors of #ac#, whose sum is #b#. Hence, here factors of #2xx2=4# are #4#and #1#, whose sum is #5#).

#(2x^2+5x+2)=(2x^2+4x+x+2)=2x(x+2)+1(x+2)=(2x+1)(x+2)#

Hence, factors of #f(x)=2x^3+x^2-8x-4# are #(x-2)(x+2)(2x+1)#.

As zeros of #f(x)# are given by #x-2=0#, #x+2=0# and #2x+1=0#

Hence, zeros of #f(x)# are #{2, -2, -1/2}#