Find the nth degree polynomial with real coefficients satisfying the given conditions

Question

Find the nth degree polynomial with real coefficients satisfying the given conditions.
n=3
f(-1)=80
zeros are 3 and 3i

The expanded and simplified polynomial is f(x)=______________

 

*****Got the answer 2x³ - 6x² + 18x - 54 multiple times and it is not correct!

Answer 1

f(x) = a(x-3)(x-3i)(x-(-3i)

 

      = a(x-3)(x-3i)(x+3i)

 

      = a(x-3)(x²+9)

 

      = a(x³+9x-3x²-27)

 

Since f(-1) = 80, we have a(-1-9-3-27) = 80

 

                                             -40a = 80

 

                                              So, a = -2

 

f(x) = -2(x³+9x-3x²-27) = -2x³+ 6x² -18x +54

 

 

 

          

Answer 2

Updated. 

 

Two roots are given: 3 and 3i. The third root must be -3i.

 

The form of the factorized polynomial is 

 

f(x) = a (x-3) (x-3i) (x+3i)

 

Applying f(-1) = 80

 

80 = a ((-1) - 3) ((-1) - 3i) ((-1) + 3i)

 

a = -2

 

Not sure why your answer is listed as incorrect.