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.

B? ? i'm most likely wrong but i think it would be b