Question

What are the zeros of f(x)= -4(2x+3)^3(6-x)
?

A
x=−32, 6, −4x=-\frac{3}{2},\ 6,\ -4x=−
2
3
​
, 6, −4

B
x=−32, 6, 0x=-\frac{3}{2},\ 6,\ 0x=−
2
3
​
, 6, 0

C
x=32, −6x=\frac{3}{2},\ -6x=
2
3
​
, −6

D
x=−32, 6x=-\frac{3}{2},\ 6x=−
2
3
​
, 6

Answer 1

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

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

This equation is satisfied if either \( (2x + 3)^3 = 0 \) or \( (6 - x) = 0 \).

1. **Finding the zeros from \( (2x + 3)^3 = 0 \):**

This gives:
\[
2x + 3 = 0
\]
\[
2x = -3
\]
\[
x = -\frac{3}{2}
\]

2. **Finding the zeros from \( (6 - x) = 0 \):**

This gives:
\[
6 - x = 0
\]
\[
x = 6
\]

So the function has zeros at:
- \( x = -\frac{3}{2} \)
- \( x = 6 \)

Now, we can summarize the zeros:
- \( -\frac{3}{2} \)
- \( 6 \)

None of these outputs represent an x-value of zero, nor do they imply any other specific value such as \( -32 \) or \( 32 \).

Now let's check the provided answer choices:

A. **\( x = -32, 6, -\frac{3}{2} \)**
B. **\( x = -32, 6, 0 \)**
C. **\( x = 32, -6, x = \frac{3}{2} \)**
D. **\( x = -32, 6, 0 \)**

None of the options give the correct zeros \( -\frac{3}{2} \) and \( 6 \).

Hence, there appears to be an inconsistency. Based on the calculated zeros for the function \( f(x) \), the correct answer should include just \( -\frac{3}{2} \) and \( 6 \).

If you need further clarification or assistance, please let me know!