If g(x)=2x4-x3+49x2-25x-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)=x2+25. Using long division,
divide x2+25 into the original function 2x4-x3+49x2-25x-25 to get 2x2-x-1.
Therefore the other zeros are 5i, -1/2, and 1.