Find an nth-degree polynomial function with real coefficients satisfying the given conditions calulator?

Degree = 4

Step-by-step explanation:

For the given conditions:

n = 4

i and 5i are zeros

f (-2) = 145

For zeros, it means they are a quadratic factor of the expression

It means, we will have x = ± i and x = ± 5i

therefore, the given factors are (x - i) (x + i) (x - 5i) (x + 5i)

Hence, we have the function

given degree = 4

f (x) = a (x-i) (x+i) (x-5i) (x+5i)

f (x) = a (x² + 1) (x² + 25)

Hence, substituting - 2 for x, we have

f (-2) = a (5) (29) = 145

Hence, a = 1

f (x) = x⁴ + 26x² + 25

Therefore, we can see that the given degree = 4