# Thread: Another factorising algebraic fraction

1. ## Another factorising algebraic fraction

Simplify:
(x^2 - y^2 - 2x - 2y) X (xy-3y^2)
(x^2 - 2xy - 3y^2) X (y-x+2)
Guide through it will be appreciated, thanks

2. $\displaystyle \frac{(x^2 - y^2 - 2x - 2y)(xy - 3y^2)}{(x^2 - 2xy - 3y^2)(y - x + 2)}$

Take out a factor of y on the top from the last bracket, and factorise the first bracket on the bottom:

$\displaystyle = \frac{y(x^2 - y^2 - 2x - 2y)(x - 3y)}{(x - 3y)(x + y)(y - x + 2)}$

Cancel the $\displaystyle (x - 3y)$ term which appears on both the top and bottom:

$\displaystyle = \frac{y(x^2 - y^2 - 2x - 2y)}{(x + y)(y - x + 2)}$

Expand the brackets on the bottom:

$\displaystyle = \frac{y(x^2 - y^2 - 2x - 2y)}{-x^2 + y^2 + 2x + 2y}$

Take out a factor of $\displaystyle -1$ on the bottom:

$\displaystyle = \frac{y(x^2 - y^2 - 2x - 2y)}{-(x^2 - y^2 - 2x - 2y)}$

Now cancel the term $\displaystyle (x^2 - y^2 - 2x - 2y)$ which appears on the top and bottom of the fraction:

$\displaystyle = \frac{y}{-1} = -y$

3. Thanks!

4. Hello, Supaweak!

Another approach . . .

Simplify: .$\displaystyle \frac{(x^2-y^2-2x-2y)\cdot(xy-3y^2)}{(x^2-2xy-3y^2)\cdot(y-x+2)}$

Factor each expression . . .

. . $\displaystyle x^2-y^2 - 2x - 2y \:=\:(x-y)(x+y) - 2(x+y) \:=\:(x+y)(x-y-2)$

. . $\displaystyle xy-3y^2 \:=\:y(x-3y)$

. . $\displaystyle x^2-2xy - 3y^2 \:=\:(x+t)(x-3y)$

. . $\displaystyle y - x + 2 \:=\:-(x-y-2)$

The problem becomes: .$\displaystyle \frac{(x+y)(x-y-2)\cdot y(x-3y)}{(x+y)(x-3y) \cdot -(x-y-2)}$

Reduce: . $\displaystyle \frac{{\color{blue}\rlap{//////}}(x+y){\color{red}\rlap{/////////}}(x-y-2)\cdot y{\color{green}\rlap{///////}}(x-3y)}{{\color{blue}\rlap{//////}}(x+y){\color{green}\rlap{///////}}(x-3y) \cdot -{\color{red}\rlap{/////////}}(x-y-2)} \;=\;\frac{y}{\text{-}1} \;=\;\boxed{-y}$

5. Thanks Soroban! I was gonna ask how (x-y)(x+y) - 2(x+y) = (x+y)(x-y-2) worked but I get it now, Thanks again for your help!