Help me factor this?

Question

(x^2-4)/(x^3+2x^2+x+2)
I also need domain and vert asymptote, please

Thanks I think I got it. I didn't need to simplify the sqr. just leave with its factors and figure out all the values for which the denom. = 0 , for the domain and asympt.

Answer 1

(x - 2) (x+ 2)
-----------------
(x^2 +1)(x+2)


(The domain is all Reals except -2)

The vertical assymptote comes at -2

Answer 2

(x^2-4)/ (x^3+2x2+x+2). When factoring, start with a divisor that divides the coefficient of both the highest order and lowest.

So, for the denominator, I choose (x+2) I chose the plus since all the terms are positive.

The numerator uses the difference of two squares (x+2)(x-2). Whenever there is only two terms in a expression, and the units term is a perfect square, and the order of the first term is even, you can play this game.

Doing that, I get (x+2)(x-2)/(x+2)(x^2+1) Simplifying the expression:

(x-2)/(x^2+1) I suggest you use Google for the domain and vertical asymptote. I don't know the answer

Answer 3

using synthetic division, with i am some what rusty with, i obtained
(x^2-4)=(x+2)(x-2)
with the rest i used the value 2 in the dvision,
2| 1 2 1 2
|__2_0_2_
1 0 1 0
this mean that (x+2)(x^2+1) is what is left
x^2+1 will give you 2 imaginary roots, they are +/- 1i

I am very sorry about the asymotote and the domain, but it should be easy from here

Answer 4

(x^3 +2x^2 +x +2)
(x^2+1)(x+2)
x^2 +1 = 0
x = i (no root)
x+2 = 0
x = -2
therefore
D: x \= -2
vert asymptote is at x = -2