If g(x)=2x^{4}-x^{3}+49x^{2}-25x-25
and one of the zeros of g(x) is -5i, what are all the other

three zeros of the function (in addition to -5i)?

Because we know that -5i is a zero and the polynomial has to have real coefficients,

we know that the conjugate 5i must also be a zero.

Multiplying
these two known factors (x-5i)(x+5i)=x^{2}+25. Using long division,

divide x^{2}+25
into the original function 2x^{4}-x^{3}+49x^{2}-25x-25
to get 2x^{2}-x-1.

2x^{4}-x^{3}+49x^{2}-25x-25=(x^{2}+25)(2x^{2}-x-1)=(x-5i)(x+5i)(2x+1)(x-1)

Therefore the other zeros are 5i, -1/2, and 1.