anonymous
  • anonymous
Algebra equation! simplify completely (x^2 + x - 12) / (x^2 - x - 20) divided by (3x^2 -24x + 45) / (12x^2 - 48x - 60)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
@Luigi0210 @zepdrix please help
zepdrix
  • zepdrix
\[\large\rm \frac{x^2 + x - 12}{x^2 - x - 20}\div\frac{3x^2 -24x + 45}{12x^2 - 48x - 60}\]We gotta simplify huh? :) So I guess we'll need to do some good ole factoring.

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anonymous
  • anonymous
1 ——————————— 36 • (x - 5)^3• (x + 1)
anonymous
  • anonymous
Well, dont we first flip the second fraction and switch the sign to multiplication? I just cant remember if we change the signs within the fraction when we flip it..
anonymous
  • anonymous
thats what i got but could be wrong
zepdrix
  • zepdrix
Yes, flip is a good first step!
anonymous
  • anonymous
@yinkim52001 Thanks so much for trying to help! But I want to learn how to do this as I am struggling with the problem (:
zepdrix
  • zepdrix
\[\large\rm \frac{x^2 + x - 12}{x^2 - x - 20}\cdot\frac{12x^2 - 48x - 60}{3x^2 -24x + 45}\]
anonymous
  • anonymous
So the signs dont change within the fraction?
zepdrix
  • zepdrix
correct. no change. example:\[\large\rm \frac{\left(\frac{a}{b}\right)}{\left(\frac{c}{d}\right)}=\frac{a}{b}\cdot\frac{d}{c}\]This is how we apply this "flip" idea ^ See how nothing changed to negative? :o
anonymous
  • anonymous
Yes! So what is the next step? I factor both denominators and both numerators?
zepdrix
  • zepdrix
Yesss. Stuck on any of those steps? :)
anonymous
  • anonymous
Not yet! Lol I have to try it out. Will you please give me a minute?
zepdrix
  • zepdrix
k :)
anonymous
  • anonymous
Okay! So I have (x-3)(x+4) / (x-4)(x+5) times 12(x^2 - 4x -5) / 3(x^2 - 8x + 15)
anonymous
  • anonymous
I'm not sure how to further factor the right fraction because the top has a prime number (5) and the denominator is giving me an issue too..
zepdrix
  • zepdrix
\[\large\rm \frac{(x-3)(x+4)}{\color{red}{(x-4)(x+5)}}\cdot\frac{12(x^2-4x-5)}{3(x^2-8x+15)}\]Woops! :O Double check the red a sec. Rest looks good though.
anonymous
  • anonymous
I put that ;p
zepdrix
  • zepdrix
ya it's incorrect! :O fix it silly!
zepdrix
  • zepdrix
The middle term is negative, so the LARGER number should get the negative sign, ya? :)
anonymous
  • anonymous
I guess it would have to be (x + 4) and (x - 5) haha
zepdrix
  • zepdrix
\[\large\rm \frac{(x-3)(x+4)}{(x+4)(x-5)}\cdot\frac{12(x^2-4x-5)}{3(x^2-8x+15)}\]Ooooo kay great!
anonymous
  • anonymous
Yes (:
zepdrix
  • zepdrix
You are correct about the 5 being prime. Which means, IF IT CAN FACTOR, the only options we have will be 1 and 5, ya?
anonymous
  • anonymous
Ohhh! Right, I didn't even think about that for some reason ;p Let me try to factor further!
anonymous
  • anonymous
Okay so that would factor out to (x - 5) (x + 1), but I tried factoring out the denominator and I couldn't figure it out because neither (x + 5)(x - 3) nor (x + 3)(x - 5) were foiling out to be (x^2 - 8x + 15)..
anonymous
  • anonymous
WOW!
anonymous
  • anonymous
Just realized they could noth be negative... Okat my bad
anonymous
  • anonymous
both* okay*
zepdrix
  • zepdrix
ya there ya go :) same sign since the 15 is positive
anonymous
  • anonymous
Right! So let me see what I can do this far & I'll come back with my answer or if I get stuck ;p
zepdrix
  • zepdrix
k brb gonna make a sammich c:
anonymous
  • anonymous
No problem ;p
anonymous
  • anonymous
Got it!!! It's 4(x+1) / x - 5 Thank you so so much! <3
zepdrix
  • zepdrix
yayyy good job \c:/
anonymous
  • anonymous
Thaaank you (:

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