Which factorization can be used to reveal the zeros of the function

f(n) = -12n² - 11n + 15?

A.) f(n) = -n(12n + 11) + 15

B.) f(n) = (-4n + 3)(3n + 5)

C.) f(n) = -(4n + 3)(3n + 5)

D.) f(n) = (4n + 3)(-3n + 5)

f(n) = -12n² - 11n + 15?

A.) f(n) = -n(12n + 11) + 15

B.) f(n) = (-4n + 3)(3n + 5)

C.) f(n) = -(4n + 3)(3n + 5)

D.) f(n) = (4n + 3)(-3n + 5)

f(n) = -12n² - 11n + 15

= -1(12n² + 11n - 15) {factored -1 out}

12n² + 11n - 15 needs to be factored by grouping:

- split the middle term, 11n, into two terms who's coefficients

multiply to get what 12(-15) is......-180

and add to get the middle coefficient 11

So, two numbers that multiply to get -180 and add to get 11 are.......20 and -9

12n² + 11n - 15

= 12n² + 20n - 9n - 15 {split the middle term, 11n, into 20n and -9n}

= 4n(3n + 5) - 3(3n + 5) {factoring by grouping, factored 4n out of 1st two terms and -3 out of last two}

= (4n - 3)(3n + 5) {factored out the common factor, (3n + 5)}

f(n) = -1(12n² - 11n + 15)

f(n) = -1(4n - 3)(3n + 5) {factored by grouping}

so that it would match an answer choice}

= -1(12n² + 11n - 15) {factored -1 out}

12n² + 11n - 15 needs to be factored by grouping:

- split the middle term, 11n, into two terms who's coefficients

multiply to get what 12(-15) is......-180

and add to get the middle coefficient 11

So, two numbers that multiply to get -180 and add to get 11 are.......20 and -9

12n² + 11n - 15

= 12n² + 20n - 9n - 15 {split the middle term, 11n, into 20n and -9n}

= 4n(3n + 5) - 3(3n + 5) {factoring by grouping, factored 4n out of 1st two terms and -3 out of last two}

= (4n - 3)(3n + 5) {factored out the common factor, (3n + 5)}

f(n) = -1(12n² - 11n + 15)

f(n) = -1(4n - 3)(3n + 5) {factored by grouping}

**B.) f(n) = (-4n + 3)(3n + 5)**{distributed the negative sign through 1st set,so that it would match an answer choice}

*Ask Algebra House*