Conjugate roots theorem

Question

Given that 3 – 5i is a zero, factor the following polynomial function completely. Use the Conjugate Roots Theorem, if applicable. F(x)=x^4-2x^3-2x^2+208x-408

Answer 1

Using the conjugate of the root, one factor of F(x) is

 

(x - [3 - 5i])(x - [3 + 5i]) =

 

x² - x(3 + 5i) - x(3 - 5i) + (9 + 25) =

 

x² - 6x + 34

 

 

Next, divide F(x) by    (x² - 6x + 34) using long division.

 

 

                                          x²     + 4x    - 12

                      ____________________________

(x² - 6x + 34) | x⁴ - 2x³ -   2x²    + 208x - 408

                       x⁴ - 6x³ +34x²

                       ____________________

                            4x³  - 36x²    + 208x

                            4x³ - 24x²     + 136x

                            _______________________

                                  -12x²      +  72x   - 408

                                  -12x²      +  72x   - 408

                                _______________________

                                                                0

 

 

 

F(x) = (x² - 6x + 34)(x² + 4x - 12)

F(x) = (x² - 6x + 34)(x - 2)(x + 6)