Question

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

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:

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

Since the product of the factors equals zero, we can set each factor equal to zero individually.

1. **First factor:** \( (2x + 3)^3 = 0 \)

To solve this, we set the base equal to zero:

\[
2x + 3 = 0
\]

Subtracting 3 from both sides gives:

\[
2x = -3
\]

Dividing by 2, we find:

\[
x = -\frac{3}{2}
\]

2. **Second factor:** \( 6 - x = 0 \)

Solving this gives:

\[
x = 6
\]

Now we can summarize the zeros of the function \( f(x) \):

- From \( (2x + 3)^3 = 0 \), we have a triple zero at \( x = -\frac{3}{2} \).
- From \( 6 - x = 0 \), we have a single zero at \( x = 6 \).

Thus, the zeros of the function are:

\[
x = -\frac{3}{2} \quad (\text{triple root}) \quad \text{and} \quad x = 6 \quad (\text{single root})
\]