Question

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

Answer 1

To find the number of zeros of the function

\[
f(x) = x(x-1)(2x+4)^2,
\]

we can identify the zeros by setting \( f(x) = 0 \) and solving for \( x \).

The function is a product of three factors:

1. \( x = 0 \)
2. \( x - 1 = 0 \) which gives \( x = 1 \)
3. \( (2x + 4)^2 = 0 \)

Now we solve the third factor:

\[
2x + 4 = 0 \quad \Rightarrow \quad 2x = -4 \quad \Rightarrow \quad x = -2.
\]

Since \( (2x + 4)^2 \) is squared, the zero at \( x = -2 \) has multiplicity 2.

Now we can summarize the zeros along with their multiplicities:

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

To find the total number of zeros, including their multiplicities, we add them up:

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

Thus, the total number of zeros, including repeated zeros, is

\[
\boxed{4}.
\]