To determine which factorizations can be used to reveal the zeros of the function [tex]\( f(n) = -12n^2 - 11n + 15 \)[/tex], we need to analyze each given option carefully.
### Option A:
[tex]\[ f(n) = -n(12n + 11) + 15 \][/tex]
Let's rewrite and expand this option to see if it matches the original function.
[tex]\[ -n(12n + 11) + 15 = -12n^2 - 11n + 15 \][/tex]
This matches the original function exactly. However, this form does not directly provide a factorization suitable for finding zeros because it does not clearly separate the polynomial into product terms that can be set to zero.
### Option B:
[tex]\[ f(n) = (-4n + 3)(3n + 5) \][/tex]
Let's expand this factorization.
[tex]\[ (-4n + 3)(3n + 5) = (-4n \cdot 3n) + (-4n \cdot 5) + (3 \cdot 3n) + (3 \cdot 5) \][/tex]
[tex]\[ = -12n^2 - 20n + 9n + 15 \][/tex]
[tex]\[ = -12n^2 - 11n + 15 \][/tex]
This matches the original function as well. Since this is a factorization into the product of two binomials, it can be directly used to find the zeros of the function by solving:
[tex]\[ (-4n + 3) = 0 \text{ or } (3n + 5) = 0 \][/tex]
[tex]\[ n = \frac{3}{4} \text{ or } n = -\frac{5}{3} \][/tex]
### Option C:
[tex]\[ f(n) = -(4n + 3)(3n + 5) \][/tex]
Let's expand this factorization.
[tex]\[ -(4n + 3)(3n + 5) = -(4n \cdot 3n + 4n \cdot 5 + 3 \cdot 3n + 3 \cdot 5) \][/tex]
[tex]\[ = -(12n^2 + 20n + 9n + 15) \][/tex]
[tex]\[ = -12n^2 - 29n - 15 \][/tex]
This does not match the original function, so it is not a suitable factorization.
### Option D:
[tex]\[ f(n) = (4n + 3)(-3n + 5) \][/tex]
Let's expand this factorization.
[tex]\[ (4n + 3)(-3n + 5) = (4n \cdot -3n) + (4n \cdot 5) + (3 \cdot -3n) + (3 \cdot 5) \][/tex]
[tex]\[ = -12n^2 + 20n - 9n + 15 \][/tex]
[tex]\[ = -12n^2 + 11n + 15 \][/tex]
This does not match the original function, so it is not a suitable factorization.
### Conclusion:
The only factorizations that match the original function [tex]\( f(n) = -12n^2 - 11n + 15 \)[/tex] and can be used to find its zeros are:
- Option B: [tex]\( f(n) = (-4n + 3)(3n + 5) \)[/tex]
