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Question 27997: i am suppossed to simplify this rational expression. i did a whole page of them just fine but this one has had me stuck for over an hour. can you please help me 6x^3y^2-51x^2y^2-9xy^2 over 4x^4y+4x^3y+x^2y
i factored a 3xy^2out of the top and a x^2y out of the bottom. i know i should have 3xy^2(2x^2-17x-3)over x^2y(4x^2+4x+1) then i need to end up with something like (2x-?)(x+?) over (?+?)(?+?)
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! [6x^3y^2-51x^2y^2-9xy^2]/ [4x^4y+4x^3y+x^2y]
nr = [6x^3y^2-51x^2y^2-9xy^2]-----(1)
=3xy^2[2x^2-17x-3]
dr=[4x^4y+4x^3y+x^2y]
=x^2y(4x^2+4x+1)
=x^2y(2x+1)^2 ----(2)
Your arguments are right and you have proceeded quite mathematically!
Observing your requirements and the nr and dr after simplification as far as they went,the quantity -51 is the culprit and it should have been just -15
(may be that 1 that takes the unit's place should have been in the tenth place and vice versa for 5, that is 15 is inadvertantly entered as 51)
With -15 in the place of -51, the problem is quite simple and comes under simplifiable category.
Given [6x^3y^2-15x^2y^2-9xy^2]/ [4x^4y+4x^3y+x^2y] say
= 3xy^2[2x^2-5x-3]/x^2y(4x^2+4x+1)
=(3y/x)[(2x+1)(x-3)]/[(2x+1)(2x+1)]
(cancelling (xy) in the nr and in the dr
and factorising the nr and dr)
=3y(x-3)/x(2x+1) (cancelling (2x+1))
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