We know that if (x+a) is a zero of a function f(x) then f(-a)=0.
Given is f(x)=x^3 - 3x^2 -15x +125
Let us choose one factor as (x+5). This is trial and error, notice that there is 125 as the last term therefore this is a possible factor.
Now f(-5)=(-5)^3 - 3(-5)^2 -15(-5) +125
or f(-5)=-125 - 75 +75 +125=0
Hence (x+5) is a factor.
Writing the given equation as a multiplication of the factor and a quadratic in general form
(x+5)(ax^2 + bx + c) = x^3 - 3x^2 -15x +125
a x^3 + bx^2 + cx + 5ax^2 + 5bx + 5c = x^3 - 3x^2 -15x +125
Put like terms together
ax^3 + bx^2 + 5ax^2 + cx + 5bx + 5c = x^3 - 3x^2 -15x +125
ax^3 + x^2(b+5a) + x(c+5b) + 5c = x^3 - 3x^2 -15x +125
Compare coefficients of like terms
a=1
b + 5a = -3 , Inserting value of a gives us b=-8
c + 5b = -15, Inserting value of b gives us c=25
5c = 125 , Gives us again c=25
Now rewrite polynomial as
f(x)=(x+5)(x^2 -8x +25)
To find remaining two zeros using quadratic formula we obtain
x={-b+-sqrt{b^2-4ac}}/{2a}
x={-(-8)+-sqrt{(-8)^2-4xx1xx25}}/{2xx1}
x={8+-sqrt{64-100}}/2
x={8+-sqrt{-36}}/2
x={8+-6i}/2
x=4+-3i
