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\[(x+1)^{2}/x^2+2x-3 \times x^2-2x+1/x^2+4x+3\]
|dw:1341370090447:dw|
\(\huge\text{Is the problem: }\frac{(x + 1)^{2}}{x^{2} + 2x - 3} \times \frac{x^{2} - 2x + 1}{x^{2} + 2x + 3}?\)

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Other answers:

yes, i didnt know how to type it in the equation like that
You have to use the equation editor with a few extra steps. I assume you know how to factor the three trinomials?
Do you????
im not sure
here's what ive done so far
can you please tell me if it's somewhat right
Sure.
|dw:1341370527396:dw|
oh that's wrong, i factor right ?
Um...before you do that, factor the three individual trinomials. Try factoring: \(x^{2} + 2x - 3\).
\[(x-2)(x+1)\]
is that right?
Not quite. What are factors of -3 that add to 2?
-2,-1
No. -2 x -1 = 2. Not -3. You did it the opposite way. They should add to 2 and multiply to -3.
OHH ! okay
-1,3
YAY! \(x^{2} + 2x - 3 = (x - 1)(x + 3)\) Now, let's factor \(x^{2} - 2x + 1\)
-1,-1
(x-1)(x-1)
Yup. Now finally, let's factor x^{2} + 4x + 3
(x+3)(x+1)
|dw:1341371332480:dw|
|dw:1341371477936:dw|
You're getting really good! \(\large\frac{(x + 1)(x + 1)}{(x + 3)(x - 1)} \times \frac{(x - 1)(x - 1)}{(x + 3)(x + 1)} = \frac{\cancel{(x + 1)}(x + 1)}{(x + 3)\cancel{(x - 1)}} \times \frac{\cancel{(x - 1)}(x - 1)}{(x + 3)\cancel{(x + 1)}} = \frac{(x + 1)(x - 1)}{(x + 3)(x + 3)}\)Yup :) You got it!
Thankyou so much ! :)
so this is the farthest it ca be simplified right

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