Find all real zeros and complexof each equation f(x)=(x^2+4)(x^2-9)

x = ± 3, x = ± 2i

Step-by-step explanation:

Given

f(x) = (x² + 4)(x² - 9)

To find the zeros let f(x) = 0, that is

(x² + 4)(x² - 9) = 0

Equate each factor to zero and solve for x

x² - 9 = 0 ( add 9 to both sides )

x² = 9 ( take the square root of both sides )

x = ± = ± 3

x² + 4 = 0 ( subtract 4 from both sides )

x² = - 4 ( take the square root of both sides )

x = ± = ± = ± × = ± 2i

Thus zeros are

x = - 3, x = 3 ← real

x = - 2i, x = 2i ← complex

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong