n=4 f(1)=-234 zeros 3 and 4i

Question

find an nth degree polynomial

Answer 1

Your zeros are

 

x₁ = 3

x₂ = -4i

x₃ = 4i

 

And since you have a 4th degree polynomial, you need more root to have 4 zeros.  Supposed the first root has a multiplicity of 2.  This is because conjugate complex roots are associated with 2nd degree polynomials.  -4i is the conjugate of 4i.

 

f(x) = C(x - 3)²(x + 4i)(x - 4i)

 

where C is a constant.  This C will be found using the condition f(1)=-234

 

f(x) = C(x - 3)²(x² + 16)

 

 

Now we use the initial value condition to solve for C.

 

-234 = C(1 - 3)²(1² + 16)

-234 = C(4)(17)

-234 = 68C

 

-234/68 = C

 

f(x) = (-234/68)(x - 3)(x - 3)(x² + 16)

Answer 2

N=4 means a 4th degree polynomial that would have 4 roots.  Since 4i is a given root then -4i must also be a root.  Let the fourth root = a.  We can set up the polynomial by multiplying the 4 corresponding factors:

 

f(x) = (x-3)(x-a)(x+4i)(x-4i)  and simplify to:

 

f(x) = (x-3)(x-a)(x²+16)  since f(1) = -234, we can solve for a

 

-234 = (1-3)(1-a)(1²+16)

 

-234=(-2)(1-a)(17)

 

117 = 17-17a

 

100 = -17a

 

a = -100/17  so our final polynomial becomes:

 

(x-3)(x+100/17)(x²+16)