Hi, how do I fully solve this problem?
I've factorised both denominators, but not the numerators.
Please help.
Thanks
You have the following:
. . . . .$\displaystyle \frac{2x^2\, -\, 3xy\, +\, 4x\, -\, 6y}{8x^3\, -\, 27y}\, \times\, \frac{4x^2\, +\, 6xy\, +\, 9y}{x^2\, +\, x\, -\, 2}$
You say that you have factored both of the denominators. I do not see any way of factoring the first denominator, since it is not a difference of squares or cubes, and contains no common factors, so I would be interested in seeing your work for this. I will guess that the other factorization led to the following:
. . . . .$\displaystyle \frac{2x^2\, -\, 3xy\, +\, 4x\, -\, 6y}{8x^3\, -\, 27y}\, \times\, \frac{4x^2\, +\, 6xy\, +\, 9y}{(x\, +\, 2)(x\, -\, 1)}$
To learn how to factor the first numerator, try here. You should get the following:
. . . . .$\displaystyle \frac{(2x\, -\, 3y)(x\, +\, 2)}{8x^3\, -\, 27y}\, \times\, \frac{4x^2\, +\, 6xy\, +\, 9y}{(x\, +\, 2)(x\, -\, 1)}$
Since the last term in the second numerator is not $\displaystyle 9y^2$, I don't see any factoring you can do here, either. So you're stuck with just the one cancellation, and then multiplying the rest out.