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
= 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 Comments are closed.

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