Question

in -13^2(5x+10)(x-3), (x-3) has the zero of positive three, how do I find the other two zeroes

Answer 1

I want to be sure I read your expression correctly. There are two likely meanings; I'll give both.

1) If you meant -13^2(5x+10)(x-3) (i.e. -169·(5x+10)(x-3)):
-169 ≠ 0, so zeros come from the factors (5x+10)=0 and (x-3)=0.
Solve 5x+10=0 → x = -2, and x-3=0 → x = 3.
So the zeros are x = -2 and x = 3.

2) If you meant -13 x^2 (5x+10)(x-3) (i.e. -13·x^2·(5x+10)·(x-3)):
Set each factor equal to zero.
x^2 = 0 → x = 0 (a double root),
5x+10 = 0 → x = -2,
x-3 = 0 → x = 3.
So the zeros are x = 0 (multiplicity 2), x = -2, and x = 3.

Which version did you mean?