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Question 502904: Simplify the rational expression:
(7x^2y+21xy^2-28y^3)
divided by
(36x^3y-27x^2y^2-9xy^3)
So far this is what I've done:
I took a 7y out of the numerator leaving me with 7y (x^2+3xy-4y^2)
then I took a 9xy out of the denominator leaving me with 9xy (4x^2-3xy-y^2)
I now do not know what to do next because I've tried factoring out the top and bottom and keep getting stumped because things dont seem to add up right etc..
Answer by persian52(161)   (Show Source):
You can put this solution on YOUR website! (7x^(2)y+21xy^(2)-28y^(3))/(36x^(3)y-27x^(2)y^(2)-9xy^(3))
Factor out the GCF of 7y from each term in the polynomial.
(7y(x^(2))+7y(3xy)+7y(-4y^(2)))/(36x^(3)y-27x^(2)y^(2)-9xy^(3))
Factor out the GCF of 7y from 7x^(2)y+21xy^(2)-28y^(3).
(7y(x^(2)+3xy-4y^(2)))/(36x^(3)y-27x^(2)y^(2)-9xy^(3))
For a polynomial of the form x^(2)+bx+c, find two factors of c (-4) that add up to b (3). In this problem 4*-1=-4 and 4-1=3, so insert 4 as the right hand term of one factor and -1 as the right-hand term of the other factor.
(7y(x+4y)(x-y))/(36x^(3)y-27x^(2)y^(2)-9xy^(3))
Factor out the GCF of 9xy from each term in the polynomial.
(7y(x+4y)(x-y))/(9xy(4x^(2))+9xy(-3xy)+9xy(-y^(2)))
Factor out the GCF of 9xy from 36x^(3)y-27x^(2)y^(2)-9xy^(3).
(7y(x+4y)(x-y))/(9xy(4x^(2)-3xy-y^(2)))
Find the factors such that the product of the factors is the trinomial 4x^(2)-3xy-y^(2). This can be done by trial and error and checked using the FOIL method of simplifying polynomials.
(7y(x+4y)(x-y))/(9xy(4x+1y)(x-1y))
Reduce the expression (7y(x+4y)(x-y))/(9xy(4x+y)(x-y)) by removing a factor of y from the numerator and denominator.
(7(x+4y)(x-y))/(9x(4x+y)(x-y))
Reduce the expression by canceling out the common factor of (x-y) from the numerator and denominator.
(7(x+4y)(x-y))/(9x(4x+y)(x-y))
Reduce the expression by canceling out the common factor of (x-y) from the numerator and denominator.
(7(x+4y))/(9x(4x+y)) лл Answer.
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