como factorizar (x^3+2x^2-5x-6)/(x^3-7x+6)

Answer 1

http://en.wikipedia.org/wiki/Rational_root_theorem

http://es.wikipedia.org/wiki/Teorema_de_la_raíz_racional

You find that

x = -1, x = 2 and x = -3 are roots of x^3+2x^2-5x-6, therefore:

x^3+2x^2-5x-6 = (x+1)(x-2)(x+3)

x = 1, x = 2 and x=-3 are roots of x^3-7x+6, therefore:

x^3-7x+6 = (x-1)(x-2)(x+3)

Answer 2

how would you do it without synthetic div. can you factor it like a quadratic