How many zeros, including repeated zeros, does the function f(x)=x(x-1)(2x+4)^2 have?

To determine how many zeros the function \( f(x) = x(x - 1)(2x + 4)^2 \) has, we will find the values of \( x \) that make \( f(x) = 0 \).

The function is composed of three factors:

1. \( x \)
2. \( x - 1 \)
3. \( (2x + 4)^2 \)

We will find the zeros of each factor:

1. **For \( x = 0 \) from the first factor:**
\[
f(0) = 0
\]
This gives us a zero at \( x = 0 \).

2. **For \( x - 1 = 0 \) from the second factor:**
\[
x - 1 = 0 \implies x = 1
\]
This gives us another zero at \( x = 1 \).

3. **For \( (2x + 4)^2 = 0 \) from the third factor:**
To solve \( (2x + 4)^2 = 0 \):
\[
2x + 4 = 0 \implies 2x = -4 \implies x = -2
\]
The factor \( (2x + 4)^2 \) is squared, which means this zero \( x = -2 \) has a multiplicity of 2.

Now, we summarize the zeros and their multiplicities:

- \( x = 0 \) (multiplicity 1)
- \( x = 1 \) (multiplicity 1)
- \( x = -2 \) (multiplicity 2)

To find the total number of zeros, including repeated zeros, we add the multiplicities:

\[
1 \text{ (from } x=0\text{)} + 1 \text{ (from } x=1\text{)} + 2 \text{ (from } x=-2\text{)} = 4
\]

Thus, the function \( f(x) = x(x - 1)(2x + 4)^2 \) has a total of **4 zeros**, including repeated zeros.

The final answer is \(\boxed{4}\).