To find the number of zeros of the function \( f(x) = x(x - 1)(2x + 4)^2 \), we can set the function equal to zero and solve for \( x \).
The function can be factored as follows:
- \( x = 0 \) gives us one zero.
- \( x - 1 = 0 \) gives us another zero: \( x = 1 \).
- \( (2x + 4)^2 = 0 \) gives us the zeros from \( 2x + 4 = 0 \): \[ 2x + 4 = 0 \implies 2x = -4 \implies x = -2. \] Since this term is squared, the zero \( x = -2 \) is a repeated zero. Hence, it counts as two zeros.
Now, let's summarize the zeros:
- \( x = 0 \) (1 zero)
- \( x = 1 \) (1 zero)
- \( x = -2 \) (2 zeros, due to the square)
Putting it all together, the total count of zeros (including repeated zeros) is: \[ 1 + 1 + 2 = 4. \]
Therefore, the function \( f(x) = x(x - 1)(2x + 4)^2 \) has a total of 4 zeros, including repeats.
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