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ZeHanzBest ResponseYou've already chosen the best response.1
First, try to factor both numerator and denominator. If you're lucky, you can then divide out the common factors!
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
Can you factor y²+y20?
 one year ago

andreadesirepenBest ResponseYou've already chosen the best response.0
i woul eliminate the y^2 right?
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
No, write it as: y²+y20=(y ...)(y ...)
 one year ago

tyteen4a03Best ResponseYou've already chosen the best response.0
@andreadesirepen y^2 + y != y^3. Different powers.
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
On the dots you have to put numbers for which the sum is 1 and the product is 20. Does that sound familiar?
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
OK, then find these numbers!
 one year ago

andreadesirepenBest ResponseYou've already chosen the best response.0
is it 2x/y^2 ?
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
No, first you get: y²+y20=(y4)(y+5). That is just the numerator.
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
Now you have to do the same with the denominator: y²y12=(y ...)(y ...) On the dots are now numbers with sum 1 and product 12. What are they?
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
I get the feeling you don't understand what is going on. If I want to factor y²y12, 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 +3y4y12 = y²y12. 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.
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
So after all this factoring, we now have:\[\frac{ y^2+y20 }{ y^2y12 }=\frac{ (y+5)(y4) }{ (y+3)(y4) }\] 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 x4 as a common factor. That can therefore be divided out, leaving only:\[\frac{ y+5 }{ y+3 }\]
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
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!
 one year ago

andreadesirepenBest ResponseYou've already chosen the best response.0
it´ll take me a while to figure this. thanks though!
 one year ago

andreadesirepenBest ResponseYou've already chosen the best response.0
do you know this one ? http://learn.flvs.net/webdav/assessment_images/educator_algebra2_v10/09_14_31.gif
 one year ago

andreadesirepenBest ResponseYou've already chosen the best response.0
THIS one is tuff
 one year ago

tyteen4a03Best ResponseYou've already chosen the best response.0
@andreadesirepen You need to download the file and reupload it on OpenStudy.
 one year ago

tyteen4a03Best ResponseYou've already chosen the best response.0
@andreadesirepen Same drill. Remember that this expression can also be written as \(\frac{12x}{x^2+6x+9} \times \frac{x^29}{4x^2}\).
 one year ago

andreadesirepenBest ResponseYou've already chosen the best response.0
stay, change, flip?
 one year ago

andreadesirepenBest ResponseYou've already chosen the best response.0
or do we cross multiply?
 one year ago
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