Question

How do you find all the zeros of `x^4 − 4x^3 + 14x^2 − 4x + 13` with the zero 2-3i?

Answer 1

`x=i,-i,2+3i,2-3i`

Explanation:

Since the function has a zero of `2-3i`, and all the coefficients of the polynomial are real, you know that the complex conjugate of the zero will also be a zero.

Thus `2-3i` and `2+3i` are zeros of the polynomial.

Since `2-3i` and `2+3i` are roots, we know that two of the polynomial's factors are `(x-(2-3i))` and `(x-(2+3i))`.

`(x-(2-3i))(x-(2+3i))`

`=(x-2+3i)(x-2-3i)`

`=((x-2)+3i)((x-2)-3i)`

`=(x-2)^2-(3i)^2`

`=x^2-4x+4+9`

`=x^2-4x+13`

The remaining factors of the polynomial can be found through

`(x^4-4x^3+14x^2-4x+13)/(x^2-4x+13)=x^2+1`

Thus the remaining zeros can be solved through

`x^2+1=0`

`x^2=-1`

`x=+-i`

The function has four imaginary zeros and never crosses the `x`-axis.

graph{x^4-4x^3+14x^2-4x+13 [-5, 5, -21.47, 120]}