NED HELP. Medal and will become fan. I will post question below

Question

anonymous · Answer

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zepdrix · Answer

Hey :)$$\large\rm \frac{x}{x^2-9}-\frac{1}{x+3}=\frac{1}{4x-12}$$We would like to multiplyin both sides by the `least common multiple` of our denominators.
In order to achieve that, we need to start by factoring our denominators to see what they have in common.

zepdrix · Answer

The difference of squares will factor into the product of conjugates,$$\large\rm a^2-b^2=(a-b)(a+b)$$Do you see how we can apply this to our first denominator?$$\large\rm x^2-9\quad=x^2-3^2\quad=?$$

anonymous · Answer

no ;/

anonymous · Answer

4x -12?

zepdrix · Answer

Don't you see the rule I posted? :o
With the a's and b's? Hmm
Anyway, it will break down like this:$$\large\rm x^2-9\quad=x^2-3^2\quad=(x-3)(x+3)$$So we have:$$\large\rm \frac{x}{(x-3)(x+3)}-\frac{1}{x+3}=\frac{1}{4x-12}$$What about $\large\rm 4x-12?$
See how we can factor a 4 out of each term?