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.