Question 853003
(x^2-3x-10)(x^2-6x+13)
x^2-3x-10= (x+2) (x-5)
(x+2)(x-5)(x^2-6x+13)
(x-(3-2i)(x-(3+2i)=(x^2-6x+13)
the zeros must be factors of -130
the prime factors are -2󬊁3
There are only a few to test
-2, 2, -5 and 5 , -13 and 13