Question

If `1+3i` is a zero of `f`, what are all the zeros of `f(x)=x^4–3x^3+6x^2+2x-60`?

Answer 1

If `1+3i` is a zero of `f(x)=x^4–3x^3+6x^2+2x−60`, then so is `1-3i`. (Complex conjugates theorem) So, you should be able to obtain the quadratic that has `1+3i and 1-3i` as its zeros, then perform long division on the polynomial to find another factor and the remaining zeros!

There are two ways:

1. Expand: `(x-(1+3i))(x - (1-3i))` (factor theorem)
   `x^2-(1+3i)x-(1-3i)x+10 = x^2 - 2x + 10`

2. Use the sum and product of roots : `1+3i+1 - 3i= 2` and `(1+3i)(1-3i)= 1 + 9 = 10`. Sum is equal to `-b/a`, so `-b/a=2/1` and `c/a=10/1`. One can solve to find a = 1, b = -2 and c = 10.

Now, for the long division :

FINALLY, we can now factor `x^2-x-6=(x+2)(x-3)` and find the remaining, Real zeros, x = -2, x = 3.