Final answer:
The real zeros of the function h(x) = -3(x² + 36)(x² - 9) are x = -3 and x = 3.
Step-by-step explanation:
To find the real zeros of the function h(x) = -3(x² + 36)(x² - 9), we can set each factor equal to zero and solve for x.
Step 1: Set x² + 36 = 0 and solve for x:
x² + 36 = 0
x² = -36
x = ±√(-36)
Since the square root of a negative number is not a real number, there are no real zeros for the factor x² + 36.
Step 2: Set x² - 9 = 0 and solve for x:
x² - 9 = 0
x² = 9
x = ±√9
x = ±3
Therefore, the real zeros of the function h(x) = -3(x² + 36)(x² - 9) are x = -3 and x = 3.