Verified answer

Supposing you meant:

(x^2 – 9)^4(x^2 + 4)^3(x – 5i)^2

Use the zero product principle three times:

x^2 - 9 = 0

Factor:

(x + 3) (x - 3) = 0

Use the zero product principle twice:

x = -3

x = 3

x^2 + 4 = 0

x^2 = -4

Take the square root:

x = ± 2i

(x – 5i)^2 = 0

Take the square root:

x - 5i = 0 twice

x = 5i twice

So altogether the six zeros of P(x) are: -3, 3, 2i, -2i, 5i, and 5i