Asked by matematicas
como factorizar (x^3+2x^2-5x-6)/(x^3-7x+6)
Answers
Answered by
Count Iblis
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)
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)
Answered by
matematicas
how would you do it without synthetic div. can you factor it like a quadratic
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