Question

Using the given zero, find one other zero of f(x). Explain the process you used to find your solution.

1 - 2i is a zero of f(x) = x^4 - 2x^3 + 6x^2 - 2x + 5.

Answer 1

By the complex conjugate root theorem, the one other solution is
`1+2i`


`(x-(1-2i))(x-(1+2i))=\\ (x-1+2i)(x-1-2i)=\\ (x-1)^2-(2i)^2=\\ x^2-2x+1+4=\\ x^2-2x+5`


`(x^4-2x^3+6x^2-2x+5)/(x^2-2x+5)=x^2+1`


`x^2+1=0\\ x^2=-1\\ x=i \vee x=-i`

So, the other solutions are
`1+2i,-i,i`