Question

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

Answer 1

Answer:

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