2007-12-26 08:46:46 · 8 answers · asked by D*Star 1 in Science & Mathematics ➔ Mathematics
In order to solve you must factor the numerator and denominator:
(x^2-x-2)/(x^2-4)
=((X+1)(X-2))/((X+2)(X-2))
Then you can cancel like terms which in this case is (x-2), and simplify:
(x+1)/(x+2)
2007-12-26 08:55:11 · answer #1 · answered by Mac 2 · 1⤊ 0⤋
(x^2 - x - 2) / (x^2 - 4)
= (x - 2)(x + 1) / (x - 2)(x +2)
= (x + 1) / (x + 2)
2007-12-26 16:56:09 · answer #2 · answered by michael_p87 2 · 0⤊ 0⤋
(x-2)(x+1) / (x-2)(x+2) = (x+1)/(x+2)
2007-12-26 16:54:56 · answer #3 · answered by stanschim 7 · 0⤊ 0⤋
First factor the numerator and denominator then cancel like terms.
(x^2 - x - 2) . (x-2)(x+1)
_________ = ________ = (x+1)/(x+2)
(x^2 - 4) ...... (x-2)(x+2)
Keep in mind that you are excluding x=2 and x=-2 because that would cause the denominator to vanish and you would be dividing by zero, even though you canceled out the x-2 factor.
2007-12-26 16:57:10 · answer #4 · answered by Astral Walker 7 · 0⤊ 0⤋
x^2-x-2/ x^2-4
(x-2)(x+1) / (x-2)(x+2)
x-2 cancel
x+1/x+2
2007-12-26 20:42:18 · answer #5 · answered by Dave aka Spider Monkey 7 · 0⤊ 0⤋
x^2-x-2 factors to (x-2)(x+1)
x^2-4 factors to (x-2)(x+2)
Does that help?
2007-12-26 16:56:19 · answer #6 · answered by I ain't nothing but a hound dog. 5 · 0⤊ 0⤋
Factor out the numerator into two terms. Do the same with the denominator. You may find that they have a term in common which you can cancel.
2007-12-26 16:53:22 · answer #7 · answered by Anonymous · 0⤊ 0⤋
(x^2-x-2) / (x^2-4)
(x^2-2x+x-2) / (x+2)(x-2)
(x{x-2}+1{x-2}) / (x+2)(x-2)
({x-2}{x+1}) / (x+2)(x-2)
(x-2)(x+1) / (x+2)(x-2)
(x+1) / (x+2)
(x+1)/(x+2)
2007-12-26 16:57:28 · answer #8 · answered by Siva 5 · 0⤊ 0⤋