To find all the zeros of #P(X)=x^4−4x^3−35x^2+6x+144#, first find a factor of the independent term #144# such as

#{1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 8, -8, ....}# for which #P(x)=0#.

It is apparent that for #x=2# as well as for #x=-3#, #P(x)=0#. Hence #-2# and #3# are two zeros of #x^4−4x^3−35x^2+6x+144#

Hence, #(x-2)(x+3)# or #x^2+x-6# divides #P(x)#. Dividing latter by former, we get

#(x^4−4x^3−35x^2+6x+144)/(x^2+x-6)# = #x^2-5x-24#

As the RHS can be factorized to #(x-8)(x+3)#, #8# and #-3# are other two zeros of #x^4−4x^3−35x^2+6x+144#. But #-3# has come earlier.

Hence zeros of #x^4−4x^3−35x^2+6x+144# are #-3#, #2# and #8#.