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#
