Find all the zeros of the function. f(x)=x^2(x+3)(x^2-1).?

f(x)=x^2(x+3)(x^2-1) = x^2(x+3)(x+1)(x-1)
When f(x) =0, x = 0, -3, -1 or 1

　
f(x) = x² (x+3) (x²−1)
f(x) = x² (x+3) (x+1) (x−1)

Zeros: x = 0, −3, −1, 1

x²(x+3)(x²-1) = 0
x = 0, -3, ±1

f(x) = x^2 * (x + 3) * (x^2 -1) = 0.


It means that:


Either: x^2 = 0 ---> x = 0 (Answer).


Or: (x + 3) = 0 ---> x = -3 (Answer).


Or: (x^2 - 1) = 0 ---> x^2 = 1 ---> x = ±√1 ---> x = ±1 (Answer).

For f(x) = 0, either:

x^2 = 0
x = 0

x+3 = 0
x = -3

x^2 -1 = 0
x^2 = 1
x = 1 or -1

f(x) = x² (x + 3) (x² - 1)

When f(x) = 0
x² (x + 3) (x² - 1) = 0
x² (x + 3) (x - 1) (x + 1) = 0
x = 0 (double roots) or x = -3 or x = 1 or x = -1

I cant