Find an nth degree polynomial function satisfying the given conditions

Question

n=3 

3 and 4i are zeros 

f(2)= -60

Answer 1

For the coefficients to be real, the complex conjugate of any complex zero must also be a zero.  Hence, the 3 zeros (n = 3) are

 

3

4i

-4i

 

f(x) = a(x - 3)(x - 4i)(x + 4i) = a(x - 3)(x² + 16)

 

where a is real and a ≠ 0.

 

f(2) = -60 = a(2 - 3)(2² + 16) = -20a

 

-60 = -20a

 

a = 3

 

f(x) = 3(x - 3)(x² + 16)

 

You can expand this if you wish.  Test the answer.

 

 

Answer 2

 

 Assuming we want REAL coefficients, then -4i must also be a zero.

 

 So the polynomial is (x-3)(x - 4i)(x + 4i)

 

                               (x-3)(x^2 + 16)

 

                               x^3 - 3x^2 + 16x - 48

 

                            

The polynomial function is this polynomial multiplied by some fixed number constant.

f(2) = -60 is used to find what the constant is.

 

For some fixed number constant,

 

C ( 2^3 - 3*2^2 + 16*2 - 48) = -60

 

C ( 8 - 12 + 32 - 48) = -60

 

C ( -20) = -60

 

C = -60/-20 = 3

 

So the polynomial function is f(x) = 3( x^3 - 3x^2 + 16x - 48 )