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

n= 4;

2i and 3i are zeros;

f (-1) equals=100

f (x) = 2x⁴ + 26x² + 72

Step-by-step explanation:

Imaginary and complex roots come in conjugate pairs. So if 2i and 3i are zeros, then - 2i and - 3i are also zeros.

f (x) = a (x - 2i) (x + 2i) (x - 3i) (x + 3i)

f (x) = a (x² - 4i²) (x² - 9i²)

f (x) = a (x² + 4) (x² + 9)

f (x) = a (x⁴ + 13x² + 36)

f (-1) = 100

100 = a ((-1) ⁴ + 13 (-1) ² + 36)

100 = a (1 + 13 + 36)

100 = 50a

a = 2

f (x) = 2 (x⁴ + 13x² + 36)

f (x) = 2x⁴ + 26x² + 72