# Trying to Factorise/Simplify an Algebraic Fraction

• Dec 27th 2008, 05:37 PM
Supaweak
Trying to Factorise/Simplify an Algebraic Fraction
Hello there sorry If I've posted this in the wrong forum but I've been having trouble with this one question in a textbook and I'd be very greatful if someone could help me through it :) The question is -
Simplify:

Quote:

___(4x^2 - y^2) X (x^2 + 2xy)__
(2x^2 + 3xy - 2y^2) X (xy - 2x^2)
I Hope i've made sense it ain't the prettiest diagram but i think you can get the jist of it and just incase the X represents multiply and the ^ refer to powers. I think the base of my difficulty is when im trying to factorise and apply 2x^2 + 3xy - 2y^2 as it isnt your traditional 'ax^2 + bx + c' quadratic.

So yeah I think thats about it, thanks for reading and I hope you can help
• Dec 27th 2008, 06:09 PM
skeeter
$\displaystyle \frac{(4x^2-y^2)(x^2+2xy)}{(2x^2 + 3xy - 2y^2)(xy - 2x^2)} = \frac{(2x-y)(2x+y)(x)(x+2y)}{(2x - y)(x + 2y)(x)(y-2x)}$
• Dec 27th 2008, 06:12 PM
Supaweak
Quote:

Originally Posted by skeeter
$\displaystyle \frac{(4x^2-y^2)(x^2+2xy)}{(2x^2 + 3xy - 2y^2)(xy - 2x^2)} = \frac{(2x-y)(2x+y)(x)(x+2y)}{(2x - y)(x + 2y)(x)(y-2x)}$

Hey thanks, could you tell me how you got from

2x^2 + 3xy - 2y^2 to (2x - y)(x+2y)?

Thanks!
• Dec 27th 2008, 06:35 PM
euclid2
Quote:

Originally Posted by Supaweak
Hey thanks, could you tell me how you got from

2x^2 + 3xy - 2y^2 to (2x - y)(x+2y)?

Thanks!

The same thing done to the numerator, factoring.

$\displaystyle 2x^2+3xy-2y^2 = (2x-y)(x+2y)= 2x^2+4xy-xy-2y^2=2x^2+3xy-2y^2$
• Dec 27th 2008, 07:00 PM
Supaweak
Ahh Thanks Heaps! Instead of splitting the 3xy into 4xy and -1xy I kept splitting it into 2xy and 1xy which wasnt working out for me, thanks again!