Question

divide. (x^2+2x+1)/(x-2)/(x^2-1)/(x^2-4)

Answer 1

X=-1

Answer 2

did that help

Answer 3

yes

Answer 4

cool

Answer 5

if x=-1, you have a point of discontinuity because then part of the problem is dividing by zero which is a no-no.

I also don't think that's the goal of this problem.

Answer 6

well i did not lie X=-1

Answer 7

First, factorize the expressions.

For (x^2+2x+1), use the identity \((a+b)^2= a^2 + 2ab+ b^2\) to factorize it.
In this case, a is x, b is 1.
So, x^2 + 2x +1 = (x+1)^2

For (x^2-1), use the identity \(a^2-b^2 = (a+b)(a-b)\).
In this case, a is x, b is 1.
x^2 - 1 = (x-1)(x+1)

Similar for (x^2-4)
x^2 - 4 = x^2 - 2^2 = (x-2)(x+2)

Now, for your question,
\[\frac{\frac{(x^2+2x+1)}{(x-2)}}{\frac{(x^2-1)}{(x^2-4)}}\]\[=\frac{\frac{(x+1)^2}{(x-2)}}{\frac{(x+1)(x-1))}{(x-2)(x+2)}}\]\[=\frac{(x+1)^2}{(x-2)}\times \frac{(x-2)(x+2)}{(x+1)(x-1)}\]
Cancel the common factor to get the answer..

Answer 8

(x+1)(x-2)/x-1

Answer 9

x^2+2+1/(x-2)(x2-4)(x^2-1)

Answer 10

Hmm....Perhaps not yet finished...
(x+1)(x-2) = x^2 -2x + x -2 = x^2 - x -2

Now, use synthetic division / long division..
x
----------------
x-1 ) x^2 - x - 2
x^2 - x
---------------
-2

You can get the quotient and remainder..

Answer 11

thank you everyone good night

Answer 12

your welcome c(;

Answer 13

I just aced my test the correct answer is (x+1)(x+2)/x-1