Final answer:
The function f(x) = -2x³ + 6x² + 20x - 4 can have two zeros which are x = 3 and x = ±√10 calculated by using factorisation.
Step-by-step explanation:
To determine the number of zeros of the given function f(x) = -2x³ + 6x² + 20x - 4, we need to find the roots of the function.
The number of zeros corresponds to the number of real solutions to the equation f(x) = 0.
The function can be factored as follows: f(x) = -2x³ + 6x² + 20x - 4
= -2x²(x - 3) + 20(x - 3)
= (x - 3)(-2x² + 20)
= (x - 3)(-2)(x² - 10)
Setting f(x) equal to zero, we have: 0 = (x - 3)(-2)(x² - 10)
This equation has three factors: (x - 3), (-2), and (x² - 10). To find the zeros, we set each factor equal to zero and solve for x:
Setting (x - 3) = 0 gives x = 3.
Setting (-2) = 0 has no real solution, as -2 is a constant.
Setting (x² - 10) = 0 gives x = ±√10.
Therefore, the function f(x) has two zeros: x = 3 and x = ±√10.