# 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)]

$\frac{7y+2}{4y^2-12y-20}+\frac{8}{4y^2-24y-20}$

Not much to simplify here. And what do you mean by "use a factor of 1"?

$\frac{7y+2}{4(y^2-3y-5)}+\frac{8}{4(y^2-6y-5)}$

$\frac{7y+2}{4(y^2-3y-5)}+\frac{2}{y^2-6y-5}$

• Sep 8th 2008, 12:01 PM
Moo
Hello,

Multiplying by 1 means multiply by $\frac aa$

While you're here :
$\frac{7y+2}{4({\color{blue}y^2-3y-5})}+\frac{8}{4({\color{red}y^2-6y-5})}$, do this :

$=\frac{7y+2}{4({\color{blue}y^2-3y-5})} \cdot \frac{\color{red}y^2-6y-5}{\color{red}y^2-6y-5}+\frac{8}{4({\color{red}y^2-6y-5})} \cdot \frac{\color{blue}y^2-3y-5}{\color{blue}y^2-3y-5}$

$=\frac{(7y+2)(y^2-6y-5)+8(y^2-3y-5)}{4({\color{red}y^2-6y-5}) ({\color{blue}y^2-3y-5})}$

$=\frac{7y^3-42y^2-35y+2y^2-12y-10+8y^2-24y-40)}{4(y^2-6y-5) (y^2-3y-5)}$

$=\frac{7y^3-32y^2-71y-50}{4(y^2-6y-5) (y^2-3y-5)}$

(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 $\frac aa$

While you're here :
$\frac{7y+2}{4({\color{blue}y^2-3y-5})}+\frac{8}{4({\color{red}y^2-6y-5})}$, do this :

$=\frac{7y+2}{4({\color{blue}y^2-3y-5})} \cdot \frac{\color{red}y^2-6y-5}{\color{red}y^2-6y-5}+\frac{8}{4({\color{red}y^2-6y-5})} \cdot \frac{\color{blue}y^2-3y-5}{\color{blue}y^2-3y-5}$

$=\frac{(7y+2)(y^2-6y-5)+8(y^2-3y-5)}{4({\color{red}y^2-6y-5}) ({\color{blue}y^2-3y-5})}$

$=\frac{7y^3-42y^2-35y+2y^2-12y-10+8y^2-24y-40)}{4(y^2-6y-5) (y^2-3y-5)}$

$=\frac{7y^3-32y^2-71y-50}{4(y^2-6y-5) (y^2-3y-5)}$

(Puke) but we used the multiplication by 1 lol!

I guess I was confused about the word "simplify"(Giggle)