# Thread: Alternative method for binomial denominators?

1. ## Alternative method for binomial denominators?

I find the method that I'm being taught to change a fraction with a binomial denominator into a mixed expression quite confusing and am wondering if there is an alternative method?

For example: $\displaystyle \frac{3x^2+4xy-4y^2+2}{x+2y}$

The method I'm being shown is to place the fraction in the form of long division. I'm not sure how to write that out using Latex.

Is this the only way to approach these problems? I find it quite confusing and quite frankly a little "messy" to work in that fashion.

2. Originally Posted by ejanderson
I find the method that I'm being taught to change a fraction with a binomial denominator into a mixed expression quite confusing and am wondering if there is an alternative method?

For example: $\displaystyle \frac{3x^2+4xy-4y^2+2}{x+2y}$

The method I'm being shown is to place the fraction in the form of long division. I'm not sure how to write that out using Latex.

Is this the only way to approach these problems? I find it quite confusing and quite frankly a little "messy" to work in that fashion.
Hi ejanderson,

you could say

$\displaystyle \frac{3x^2+4xy-4y^2+2}{x+2y}=ax+by+c$

We know there are x and y parts to the result, due to the squares in the numerator.
The "c" term may be there as the division may also result in a remainder.

Hence

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

We know "a" and "b" due to the $\displaystyle x^2$ and $\displaystyle y^2$ terms.
We need to find "c".

Multiplying out...

$\displaystyle 3x^2+4xy-4y^2+2=3x^2-2xy+cx+6xy-4y^2+2cy=3x^2+4xy-4y^2+cx+2cy$

Hence

$\displaystyle 2=cx+2cy\ \Rightarrow\ c(x+2y)=2\ \Rightarrow\ c=\frac{2}{x+2y}$

Hence

$\displaystyle \frac{3x^2+4xy-4y^2+2}{x+2y}=3x-2y+\frac{2}{x+2y}$