# Simplifying rational fractions

I can't get this one either for whatever reason, spent about 20 minutes on it and I have made no progress at all.

$$\frac{x^2}{(x^2-4)} - \frac{x+1}{x+2}.$$

I know that I can simplify this into one fraction so I make it $$\frac{x^2}{x+2}-\frac{(x+1)(x^2-4)}{(x^2-4)(x+2)}$$

I then can simplify it further making the $(x^2+4)$ into $(x-2)(x+2)$ and the $(x+2)$ into $(x-1)(x+1)$ but this does not help me get the answer. I know I have to manipulate it in some counter intuitive way but I can not make it work.

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I edited the question; please check if it's still what you meant. – Davide Giraudo Dec 17 '11 at 23:13
Yes thank you, I have no idea why I could not get the fractions to work. – user138246 Dec 17 '11 at 23:15
Type '\frac{a+b}{c+d}' to get $\frac{a+b}{c+d}$. – Davide Giraudo Dec 17 '11 at 23:18
In the original expression, multiply top and bottom by $x-2$. That will make the denominator of that term $(x+2)(x-2)$, which is $x^2-4$, so you will get a common denominator. Or else, but this is more unpleasant, multiply top and bottom of first term by $x+2$, and top and bottom of second term by $x^2-4$. That will get you a common denominator of $(x+2)(x^2-4)$. – André Nicolas Dec 17 '11 at 23:18

As first step we may use the common denominator $(x^{2}-4)=(x-2)(x+2)$ because the $\text{lcm}\left( (x-2)(x+2),(x+2)\right) =(x-2)(x+2)$ $$\begin{eqnarray*} \frac{x^{2}}{(x^{2}-4)}-\frac{x+1}{x+2} &=&\frac{x^{2}}{(x-2)(x+2)}-\frac{ \left( x+1\right) (x-2)}{\left( x+2\right) (x-2)} \\ &=&\frac{x^{2}-\left( x+1\right) (x-2)}{(x-2)(x+2)}.\tag{1} \end{eqnarray*}$$

Otherwise we would get the equivalent but more more complex fraction $$\frac{x^{2}}{(x^{2}-4)}-\frac{x+1}{x+2}=\frac{x^{2}\left( x+2\right) -\left( x+1\right) (x^{2}-4)}{(x^{2}-4)\left( x+2\right) }.$$ Expanding the second term of the numerator of $(1)$ $$\begin{eqnarray*} \left( x+1\right) (x-2) &=&x(x-2)+(x-2)=x^{2}-2x+x-2 \\ &=&x^{2}-x-2 \end{eqnarray*}$$

and substituting into the fraction we get $$\frac{x^{2}-\left( x^{2}-x-2\right) }{(x-2)(x+2)}=\frac{x^{2}-x^{2}+x+2}{ (x-2)(x+2)}=\frac{x+2}{(x-2)(x+2)},\tag{2}$$ which for $x+2\neq 0$ simplifies to $$\frac{1}{x-2}\tag{3}$$

Added: In general we transform the sum (or difference) of two given rational fractions (the numerator and denominator consists of polynomials) into a single equivalent fraction, by using properties such as

1. $$\frac{A(x)}{B(x)}=\frac{A(x)P(x)}{B(x)P(x)}\qquad\text{for }P(x)\neq 0.$$
2. $$\frac{A(x)}{B(x)}\pm \frac{C(x)}{D(x)}=\frac{A(x)D(x)\pm B(x)C(x)}{B(x)D(x)}.$$
3. $$\frac{A(x)}{B_1(x)B_2(x)}\pm \frac{C(x)}{B_2(x)}=\frac{A(x)\pm B_1(x)C(x)}{B_1(x)B_2(x)}.$$
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$$\frac{x^2}{x^2-4} - \frac{x+1}{x+2} = \frac{x^2}{(x-2)(x+2)}-\frac{x+1}{x+2}=\frac{x^2}{(x-2)(x+2)}-\frac{(x-2)(x+1)}{(x-2)(x+2)}.$$

Working with the numerators: $$x^2-(x-2)(x+1) = x^2 - (x^2 -x -2).$$ Here's the easiest mistake to make (I've seen this happen zillions of times including in calculus courses):

Right: $x^2 - (x^2 -x -2) = x^2 - x^2 + x + 2$

Wrong: $x^2 - (x^2 -x -2) = x^2 - x^2 - x - 2$

The numerator ends up being $x+2$, so we get another simplification: $$\frac{x+2}{(x-2)(x+2)} = \frac{1}{x-2}.$$

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It definitely makes life easier to notice that $x^2 - 4 = (x + 2)(x - 2)$. If you multiply the second term of the original expression by $1 = \frac{x - 2}{x - 2}$ then you get $\frac{x^2}{(x + 2)(x - 2)} - \frac{(x + 1)(x - 2)}{(x + 2)(x - 2)}.$ These two fractions have a common denominator, so you can combine them into a single quotient. Work out the numerator, and see if you can cancel anything after that.

Some comments on your attempt: It isn't true that $x + 2$ equals $(x + 1)(x - 1) = x^2 - 1$. It is a good idea to look for differences of squares, though. It's certainly fine (perhaps a bit messier) to place everything over the common denominator $(x^2 - 4)(x + 2)$, but for this you would multiply the first term of your original expression by $1 = (x + 2)/(x + 2)$; could you explain how you got $x^2/(x + 2)$?

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