Question

how many zeros does the function $f(x) = x(x - 1)(2x + 4)^2$ have? (1 point)
○ 4
○ 2
○ 3
○ the degree cannot be determined from the information provided.

Explanation:

Step1: Find zeros from each factor

To find the zeros of the function \( f(x) = x(x - 1)(2x + 4)^2 \), we set each factor equal to zero.

• For the factor \( x \): \( x = 0 \) gives a zero.
• For the factor \( (x - 1) \): \( x - 1 = 0 \) implies \( x = 1 \), which is another zero.
• For the factor \( (2x + 4)^2 \): Set \( 2x + 4 = 0 \). Solving for \( x \), we get \( 2x = - 4 \), so \( x = - 2 \). Since the factor is squared, \( x = - 2 \) is a zero (with multiplicity 2, but we count the distinct zero here).

Step2: Count the distinct zeros

The distinct zeros are \( x = 0 \), \( x = 1 \), and \( x = - 2 \). Wait, no, wait: Wait, let's re - examine. Wait, the factor \( (2x + 4)^2 \) gives \( x=-2 \) (even though it has multiplicity 2, it's still one distinct zero? Wait, no, no. Wait, the question is about the number of zeros (counting multiplicity? Or distinct? Wait, in the context of polynomial zeros, when we say "how many zeros" without specifying, sometimes it can be a bit ambiguous, but in the case of factored form, let's expand the idea.

Wait, the function is a polynomial. The zeros are the values of \( x \) for which \( f(x) = 0 \). Let's find all the roots (including multiplicity? Or distinct? Wait, let's check the factors:

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

But when we count the number of zeros (in the sense of the number of solutions to \( f(x) = 0 \), counting multiplicity), we have \( 1 + 1+ 2=4 \)? Wait, no, that can't be. Wait, no, maybe I made a mistake. Wait, let's re - do:

Wait, the function is \( f(x)=x(x - 1)(2x + 4)^2 \). Let's find the roots:

1. From \( x = 0 \): root at \( x = 0 \)
2. From \( x - 1=0 \): root at \( x = 1 \)
3. From \( 2x + 4 = 0\Rightarrow x=-2 \). Since the factor is squared, \( x = - 2 \) is a root with multiplicity 2.

But when we are asked "how many zeros" the function has, if we are counting the number of roots (counting multiplicity), then the total number of roots (zeros) is \( 1+1 + 2=4 \). But wait, let's check the options. The options are 4, 2, 3, or undetermined.

Wait, maybe the question is counting distinct zeros? No, that would be 3, but 3 is an option. Wait, no, wait, I think I messed up. Wait, \( (2x + 4)^2=4(x + 2)^2 \), so the factor is \( (x + 2)^2 \). So the roots are \( x = 0 \), \( x = 1 \), and \( x=-2 \) (with multiplicity 2). But when we count the number of zeros (the number of times the graph crosses or touches the x - axis, but in terms of the number of solutions to \( f(x)=0 \), if we count multiplicity, it's 4. But let's check the options. The first option is 4.

Wait, let's expand the polynomial to check the degree. The degree of the polynomial: the degree of \( x \) is 1, degree of \( (x - 1) \) is 1, degree of \( (2x + 4)^2=4(x + 2)^2 \) is 2. So the total degree is \( 1+1 + 2=4 \). A polynomial of degree \( n \) has at most \( n \) zeros (counting multiplicity). In our case, the polynomial is of degree 4, and we have roots \( x = 0 \) (multiplicity 1), \( x = 1 \) (multiplicity 1), \( x=-2 \) (multiplicity 2). So the number of zeros (counting multiplicity) is \( 1 + 1+2 = 4 \).

Answer:

4 (the option corresponding to 4, so if the options are as given, the answer is the first option, e.g., if the first option is "4", then the answer is 4)