Question

how do I factor x^3+3x^2-6x-8?

Answer 1

(x+1)(x+4)(x-2) = 0

Answer 2

First I would re-arrange and look for patterns. Noting that x^3-8 looks pretty perfect cubes to me. using (a^3-b^3)=(a-b)(a^2+ab+b^2) we may get somewhere.

re-arrange
x^3-8+3x^2-6x

group first two and second two terms and factor
(x-2)(x^2+2x+4)+3x(x-2)

factor out x-2
(x-2)(x^2+2x+4+3x)

combine like terms
(x-2)(x^2+5x+4)

then factor the poynomial
(x-2)(x+1)(x+4)