# Thread: Simplifying and Factorising Algebraic Fractions

1. ## Simplifying and Factorising Algebraic Fractions

Hi, how do I fully solve this problem?
I've factorised both denominators, but not the numerators.

Thanks

2. Originally Posted by BG5965
Hi, how do I fully solve this problem?
I've factorised both denominators, but not the numerators.

Thanks
$\displaystyle 2x^2-3xy+4x-6y = 2x^2+4x - 6y -3xy = 2x(x+2)-3y(x+2) = (x+2)(2x-3y)$

should $\displaystyle 4x^2+6xy+9y$ be $\displaystyle y^2$ at the end?

3. Originally Posted by BG5965
I've factorised both denominators, but not the numerators.

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.