how do you factor x^3-4x^2-11x+30?

Question

how do you factor

x^3-4x^2-11x+30

Answer 1

x^3-4x^2-11x+30
=x^3+27-4x^2-11x+3
=(x^3+27)-(4x^2+11x-3)
={(x)^3+(3)^3-(4x^2+12x-x-3)
=(x+3)(x^2-3x+9)-{4x(x+3)-1(x+3)
=(x+3)(x^2-3x+9)-(x+3)(4x-1)
=(x+3)(x^2-3x+9-4x+1)
=(x+3)(x^2-7x+10)
=(x+3)(x-2)(x-5)
10

Answer 2

Given that the lead coefficient is 1, the only possible rational factors divide evenly into 30,
namely, ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30,

If none of those work, then there are no rational factors, although there still is a solution, albeit an irrational one. However, this one factors nicely.

x³ - 4x² - 11x + 30 = (x + 3)(x - 2)(x - 5)

Answer 3

You don't. Cubics don't factor easily. I would use a program, such as on a graphing calculator, that analytically calculates the roots. Or if you are a masochist, the is Cardan's Formula.