First, since we are subtracting fractions, we are going to want to find a common denominator between the terms we are subtracting. In this case, [tex] (x + 2) [/tex], [tex] (x - 2) [/tex], and [tex] x^2 [/tex] are very different terms, meaning that we are going to have to multiply both fractions by the terms they are missing in the denominators.
The term [tex] \dfrac{3x}{(x + 2)(x - 2)} [/tex] does not have an [tex] x^2 [/tex] in the denominator, meaning that we are going to need to multiply both the numerator and the denominator of the fraction by [tex] x^2 [/tex]. We have to multiply it by both the numerator and the denominator to keep the fraction similar to its prior form. Doing this results in:
[tex] \Bigg(\dfrac{3x}{(x + 2)(x - 2)}\Bigg)\Bigg(\dfrac{x^2}{x^2}\Bigg) [/tex]
[tex] \dfrac{3x^3}{(x + 2)(x - 2)(x^2)} [/tex]
In the second term which is being subtracted, the denominator is absent of the [tex] (x + 2) [/tex] and [tex] (x - 2) [/tex] terms, meaning that we will have to multiply both the numerators and denominators of the fraction by these terms to give the second fraction a like denominator:
[tex] \Bigg(\dfrac{1}{x^2}\Bigg)\Bigg(\dfrac{(x + 2)(x - 2)}{(x + 2)(x - 2)}\Bigg) [/tex]
[tex] \dfrac{(x + 2)(x - 2)}{(x + 2)(x - 2)(x^2)} [/tex]
Using these terms, our subtraction problem looks like this:
[tex] \dfrac{3x^3}{(x + 2)(x - 2)(x^2)} - \dfrac{(x + 2)(x - 2)}{(x + 2)(x - 2)(x^2)} [/tex]
We can now use our common denominator to simplify this problem to just one fraction:
[tex] \dfrac{3x^3 - (x + 2)(x - 2)}{(x + 2)(x - 2)(x^2)} [/tex]
Now, using our algebraic operations, we can simplify this fraction into something more manageable:
[tex] \dfrac{3x^3 - (x + 2)(x - 2)}{(x + 2)(x - 2)(x^2)} [/tex]
[tex] \dfrac{3x^3 - (x^2 - 4)}{(x^2 - 4)(x^2)} [/tex]
[tex] \dfrac{3x^3 - x^2 + 4}{x^4 - 4x^2} [/tex]
Our simplified expression would be:
[tex] \boxed{\dfrac{3x^3 - x^2 + 4}{x^4 - 4x^2}} [/tex]
(Keep in mind that this is the same exact expression, represented in a different way. This means that there are many, many ways to represent the original expression, but this one is one that I feel very well simplifies the original problem.)
