text: # simplify by using a factor of 1 when possible.
  
  • Sep 8th 2008, 11:16 AM
  mkpg24
  simplify by using a factor of 1 when possible.
  [(7y+2)/(4y^2-12y-20)]+[(8)/(4y^2-24y-20)]
  • Sep 8th 2008, 11:52 AM
  masters
  Quote:
  
  Originally Posted by mkpg24
  [(7y+2)/(4y^2-12y-20)]+[(8)/(4y^2-24y-20)]
  
  $\displaystyle \frac{7y+2}{4y^2-12y-20}+\frac{8}{4y^2-24y-20}$
  
  Not much to simplify here. And what do you mean by "use a factor of 1"?
  
  $\displaystyle \frac{7y+2}{4(y^2-3y-5)}+\frac{8}{4(y^2-6y-5)}$
  
  $\displaystyle \frac{7y+2}{4(y^2-3y-5)}+\frac{2}{y^2-6y-5}$
  
  • Sep 8th 2008, 12:01 PM
  Moo
  Hello,
  
  Multiplying by 1 means multiply by $\displaystyle \frac aa$
  
  While you're here :
  $\displaystyle \frac{7y+2}{4({\color{blue}y^2-3y-5})}+\frac{8}{4({\color{red}y^2-6y-5})}$, do this :
  
  $\displaystyle =\frac{7y+2}{4({\color{blue}y^2-3y-5})} \cdot \frac{\color{red}y^2-6y-5}{\color{red}y^2-6y-5}+\frac{8}{4({\color{red}y^2-6y-5})} \cdot \frac{\color{blue}y^2-3y-5}{\color{blue}y^2-3y-5}$
  
  $\displaystyle =\frac{(7y+2)(y^2-6y-5)+8(y^2-3y-5)}{4({\color{red}y^2-6y-5}) ({\color{blue}y^2-3y-5})}$
  
  $\displaystyle =\frac{7y^3-42y^2-35y+2y^2-12y-10+8y^2-24y-40)}{4(y^2-6y-5) (y^2-3y-5)}$
  
  $\displaystyle =\frac{7y^3-32y^2-71y-50}{4(y^2-6y-5) (y^2-3y-5)}$
  
  (Puke) but we used the multiplication by 1 lol!
  • Sep 8th 2008, 12:04 PM
  masters
  Quote:
  
  Originally Posted by Moo
  Hello,
  
  Multiplying by 1 means multiply by $\displaystyle \frac aa$
  
  While you're here :
  $\displaystyle \frac{7y+2}{4({\color{blue}y^2-3y-5})}+\frac{8}{4({\color{red}y^2-6y-5})}$, do this :
  
  $\displaystyle =\frac{7y+2}{4({\color{blue}y^2-3y-5})} \cdot \frac{\color{red}y^2-6y-5}{\color{red}y^2-6y-5}+\frac{8}{4({\color{red}y^2-6y-5})} \cdot \frac{\color{blue}y^2-3y-5}{\color{blue}y^2-3y-5}$
  
  $\displaystyle =\frac{(7y+2)(y^2-6y-5)+8(y^2-3y-5)}{4({\color{red}y^2-6y-5}) ({\color{blue}y^2-3y-5})}$
  
  $\displaystyle =\frac{7y^3-42y^2-35y+2y^2-12y-10+8y^2-24y-40)}{4(y^2-6y-5) (y^2-3y-5)}$
  
  $\displaystyle =\frac{7y^3-32y^2-71y-50}{4(y^2-6y-5) (y^2-3y-5)}$
  
  (Puke) but we used the multiplication by 1 lol!
  
  I guess I was confused about the word "simplify"(Giggle)
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