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This is a mix of questions
- Thread starter mcadm313
- Start date
D
Deleted member 4993
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Please share with us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
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- 1,577
\(\displaystyle Factor, \ factor, \ factor.\)
\(\displaystyle I'll \ do \ the \ first \ one \ for \ you \ to \ get \ you \ started.\)
\(\displaystyle \frac{\frac{x^2-x-2}{x^2-2x-3}}{\frac{x^2+8x+12}{x^2-3x-10}} \ = \ \frac{\frac{(x-2)(x+1)}{(x-3)(x+1)}}{\frac{(x+2)(x+6)}{(x-5)(x+2)}} \ = \ \frac{\frac{x-2}{x-3}}{\frac{x+6}{x-5}} \ = \ \frac{(x-2)(x-5)}{(x-3)(x+6)}, \ x \ \ne \ -1. -2,-6,3, \ or \ 5 \ why?\)
\(\displaystyle I'll \ do \ the \ first \ one \ for \ you \ to \ get \ you \ started.\)
\(\displaystyle \frac{\frac{x^2-x-2}{x^2-2x-3}}{\frac{x^2+8x+12}{x^2-3x-10}} \ = \ \frac{\frac{(x-2)(x+1)}{(x-3)(x+1)}}{\frac{(x+2)(x+6)}{(x-5)(x+2)}} \ = \ \frac{\frac{x-2}{x-3}}{\frac{x+6}{x-5}} \ = \ \frac{(x-2)(x-5)}{(x-3)(x+6)}, \ x \ \ne \ -1. -2,-6,3, \ or \ 5 \ why?\)