(x+1/x^2-25) * (x+5/x^2+8x+7) Please explain... x+1/(x+5)(x+7) 1/(x+5)(x+7) 1/(x-5)(x+7) 1/(x-5)(x-7)
What would x^2-25 factor to? hint: difference of squares
(x+5)(x-5)
good
what would x^2+8x+7 factor to? hint: list out all the ways to multiply to 7. Then find out which pair of factors add to 8
I already did this I just don't understand how to put it together.
tell me what x^2+8x+7 factors to
(x+1)(x+7)
yes
So \[\Large \frac{x+1}{x^2-25}*\frac{x+5}{x^2+8x+7}\] turns into \[\Large \frac{x+1}{(x+5)(x-5)}*\frac{x+5}{(x+1)(x+7)}\]
So... what does that mean?
Multiply the terms straight across to combine the fractions making \[\Large \frac{x+1}{(x+5)(x-5)}*\frac{x+5}{(x+1)(x+7)}\] turn into \[\Large \frac{(x+1)(x+5)}{(x+5)(x-5)(x+1)(x+7)}\]
Notice these (x+1) terms pair up. One in the numerator. One in the denominator \[\Large \frac{{\color{red}{(x+1)}}(x+5)}{(x+5)(x-5){\color{red}{(x+1)}}(x+7)}\]
The (x+1) terms will cancel since x/x = 1 (where x is nonzero) \[\Large \frac{{\color{red}{\cancel{(x+1)}}}(x+5)}{(x+5)(x-5){\color{red}{\cancel{(x+1)}}}(x+7)}\]
Ohh. Thanks I get it now. I forgot about the cancelling!
You also cancel (x+5)
After that cancellation of the (x+1) terms, we're left with this \[\Large \frac{x+5}{(x+5)(x-5)(x+7)}\] I'm sure you see at this point the (x+5) terms cancel as well
`You also cancel (x+5)` yes correct
So since you helped me can I give you a medal or something? I don't know if I can?
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So I did it?
yes
Okay thanks, I haven't really used openstudy.
yes @blehh10 you gave out the medal, thank you
Really, thank you!
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