Find all zeros of f(x) = x^4 - 3x^3 + 6x^2 + 2x - 60 given that 1+ 3i is a zero of f(x)?

Answer 1

Since the function has real coefficients, complex roots must appear in congugate pairs. Therefore, 1 - 3i is also a zero. By the factor theorem, [x - (1 + 3i)]*[x - (2 - 3i)] is a factor of f(x). That product is x^2 - 2x + 10 = 0. Divide f(x) by that to find f(x) = (x^2 - 2x + 10)(x^2 - x - 6) = (x^2 - 2x + 10)(x - 3)(x + 2). You now know the zeros are 1 + 3i, 1 - 3i, 3, and -2.

Answer 2

I believe its -104-46i