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

Question

n=4

i and 2i are zeros

f(-1)=20

f(x)=

Answer 1

Since i and 2i are roots, so are -i and -2i.

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

f(x) = a(x² + 1)(x² + 4)

f(-1) = 20, so, 10a = 20. So, a = 2.

f(x) = 2(x² + 1)(x² + 4) = 2(x⁴ + 5x² + 4) = 2x⁴ + 10x² + 8