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:
- \( x \)
- \( x - 1 \)
- \( (2x + 4)^2 \)
We will find the zeros of each factor:
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For \( x = 0 \) from the first factor: \[ f(0) = 0 \] This gives us a zero at \( x = 0 \).
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For \( x - 1 = 0 \) from the second factor: \[ x - 1 = 0 \implies x = 1 \] This gives us another zero at \( x = 1 \).
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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 \(\boxed{4}\).