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## andreadesirepen Group Title how would i simplify this? one year ago one year ago

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1. andreadesirepen Group Title

2. ZeHanz Group Title

First, try to factor both numerator and denominator. If you're lucky, you can then divide out the common factors!

3. ZeHanz Group Title

Can you factor y²+y-20?

4. andreadesirepen Group Title

i woul eliminate the y^2 right?

5. andreadesirepen Group Title

y^3-20

6. ZeHanz Group Title

No, write it as: y²+y-20=(y ...)(y ...)

7. tyteen4a03 Group Title

@andreadesirepen y^2 + y != y^3. Different powers.

8. ZeHanz Group Title

On the dots you have to put numbers for which the sum is 1 and the product is -20. Does that sound familiar?

9. andreadesirepen Group Title

yes

10. ZeHanz Group Title

OK, then find these numbers!

11. ZeHanz Group Title

(please)

12. andreadesirepen Group Title

is it 2x/y^2 ?

13. ZeHanz Group Title

No, first you get: y²+y-20=(y-4)(y+5). That is just the numerator.

14. ZeHanz Group Title

Now you have to do the same with the denominator: y²-y-12=(y ...)(y ...) On the dots are now numbers with sum -1 and product -12. What are they?

15. andreadesirepen Group Title

what?

16. andreadesirepen Group Title

x^2/2y?

17. ZeHanz Group Title

I get the feeling you don't understand what is going on. If I want to factor y²-y-12, which is y² -1*y -12, I must look for two numbers that have sum -1 (the coefficient of y) and product -12 (the constant at the end). These are: -4 and 3. See: -4*3 = -12, -4+3 = -1. Why does this work? Try it: (y -4)(y+3) can be expanded again, using FOIL: First: y*y = y² Outer: y*3 = 3y Inner: -4*y = -4y Last: -4*3 = -12 Together, this becomes: y² +3y-4y-12 = y²-y-12. So you have to look at it like: y² +(a+b)y + ab = (y+a)(y+b) That is why you say: the sum is the number befor the y the product is the number at the end.

18. ZeHanz Group Title

So after all this factoring, we now have:$\frac{ y^2+y-20 }{ y^2-y-12 }=\frac{ (y+5)(y-4) }{ (y+3)(y-4) }$ Only now is the moment we can simplify! Because only now we have a product in the numerator and also a product in the denominator. Both products have x-4 as a common factor. That can therefore be divided out, leaving only:$\frac{ y+5 }{ y+3 }$

19. ZeHanz Group Title

So, I'm sorry, but there is no quick way to just mess a bit with the numbers in your original fraction :( What you have to do (always!) is: 1. write numerator as a product (=factor it) 2. write denominator as a product (=factor it) 3. Look for common factors. Divide them out. 4. See what is left. This is your answer!

20. andreadesirepen Group Title

it´ll take me a while to figure this. thanks though!

21. ZeHanz Group Title

yw!

22. andreadesirepen Group Title

do you know this one ? http://learn.flvs.net/webdav/assessment_images/educator_algebra2_v10/09_14_31.gif

23. andreadesirepen Group Title

THIS one is tuff

24. tyteen4a03 Group Title

@andreadesirepen You need to download the file and reupload it on OpenStudy.

25. andreadesirepen Group Title

26. andreadesirepen Group Title

@tyteen4a03

27. tyteen4a03 Group Title

@andreadesirepen Same drill. Remember that this expression can also be written as $$\frac{12x}{x^2+6x+9} \times \frac{x^2-9}{4x^2}$$.

28. andreadesirepen Group Title

stay, change, flip?

29. andreadesirepen Group Title

or do we cross multiply?

30. andreadesirepen Group Title

@tyteen4a03