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

Apr 14, 2015

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

There are two ways:

1. Expand: $\left(x - \left(1 + 3 i\right)\right) \left(x - \left(1 - 3 i\right)\right)$ (factor theorem)
${x}^{2} - \left(1 + 3 i\right) x - \left(1 - 3 i\right) x + 10 = {x}^{2} - 2 x + 10$

2. Use the sum and product of roots : $1 + 3 i + 1 - 3 i = 2$ and $\left(1 + 3 i\right) \left(1 - 3 i\right) = 1 + 9 = 10$. Sum is equal to $- \frac{b}{a}$, so $- \frac{b}{a} = \frac{2}{1}$ and $\frac{c}{a} = \frac{10}{1}$. One can solve to find a = 1, b = -2 and c = 10.

Now, for the long division :

FINALLY, we can now factor ${x}^{2} - x - 6 = \left(x + 2\right) \left(x - 3\right)$ and find the remaining, Real zeros, x = -2, x = 3.