Which correctly factored form of the function f(x) = 36x² + 15x - 6 can be used to identify the zeros?

A.) f(x) = (4x - 1)(3x + 2)

B.) f(x) = (12x - 2)(3x + 3)

C.) f(x) = 3(4x - 1)(3x + 2)

D.) f(x) = 3(12x - 2)(3x + 3)

A.) f(x) = (4x - 1)(3x + 2)

B.) f(x) = (12x - 2)(3x + 3)

C.) f(x) = 3(4x - 1)(3x + 2)

D.) f(x) = 3(12x - 2)(3x + 3)

Look for a common factor to factor out, first.

36x² + 15x - 6

= 3(12x² + 5x - 2) {factored 3 out}

Factor the remaining quadratic trinomial by grouping.

12x² + 5x - 2

= 12x² + 8x - 3x - 2 {split the 5x into two terms, whose coefficients multiply to get 12(-2) and add to get 5}

= 4x(3x + 2) - 1(3x + 2) {factored 4x out of first two terms and -1 out of last two terms}

= (4x - 1)(3x + 2) {factored (3x + 2) out of the two terms}

Don't forget the 3 that was factored out earlier.

36x² + 15x - 6

= 3(12x² + 5x - 2) {factored 3 out}

Factor the remaining quadratic trinomial by grouping.

12x² + 5x - 2

= 12x² + 8x - 3x - 2 {split the 5x into two terms, whose coefficients multiply to get 12(-2) and add to get 5}

= 4x(3x + 2) - 1(3x + 2) {factored 4x out of first two terms and -1 out of last two terms}

= (4x - 1)(3x + 2) {factored (3x + 2) out of the two terms}

Don't forget the 3 that was factored out earlier.

**C.) f(x) = 3(4x - 1)(3x + 2)**

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