What are the zeros of f(x)=−4(2x+3)3(x−6)

(1 point)
Responses

x=−32, 6
x=−32, 6

x=32, −6
x=32, −6

x=−32, 6, −4
x=−32, 6, −4

x=−32, 6, 0

User Icon for Bot GPT 4-o mini Bot GPT 4-o mini answered
1 hour ago

To find the zeros of the function \( f(x) = -4(2x + 3)^3(x - 6) \), we need to set the function equal to zero and solve for \( x \):

\[ -4(2x + 3)^3(x - 6) = 0 \]

Since the product is equal to zero when any factor is equal to zero, we can set each factor to zero:

  1. For \( (2x + 3)^3 = 0 \):

    • Taking the cube root, we have \( 2x + 3 = 0 \): \[ 2x = -3 \quad \Rightarrow \quad x = -\frac{3}{2} = -1.5 \]
  2. For \( (x - 6) = 0 \): \[ x = 6 \]

Thus, the zeros of the function are \( x = -\frac{3}{2} \) and \( x = 6 \).

Now, looking at your provided options:

  1. \( x = -32, 6 \)
  2. \( x = 32, -6 \)
  3. \( x = -32, 6, -4 \)
  4. \( x = -32, 6, 0 \)

None of these options contain the correct zeros. The calculated zeros are \( x = -1.5 \) (or \( -\frac{3}{2} \)) and \( x = 6 \). Since you did not provide an exact match in your response options, check your given options more carefully.