What is the completely factored form of f (x) = 6x^3-13x^2-4x+15?

a. (x+1) (6x^2-19x+15)

b. (x+1) ^2 (2x-3)

c. (x+1) (3x-2) (5x-3)

d. (x+1) (2x-3) (3x-5)

F (x) = 6x³ - 13x² - 4x + 15

f (x) = 6x³ - 19x² + 6x² + 15x - 19x + 15

f (x) = 6x² - 19x² + 15x + 6x² - 19x + 15

f (x) = x (6x²) - x (19x) + x (15) + 1 (6x²) - 1 (19x) + 1 (15)

f (x) = x (6x² - 19x + 15) + 1 (6x² - 19x + 15)

f (x) = (x + 1) (6x² - 19x + 15)

f (x) = (x + 1) (6x² - 10x - 9x + 15)

f (x) = (x + 1) (2x (3x) - 2x (5) - 3 (3x) + 3 (5))

f (x) = (x + 1) (2x (3x - 5) - 3 (3x - 5))

f (x) = (x + 1) (2x - 3) (3x - 5)

The answer is D.