• Jun 25th 2013, 10:36 AM
fogownz
Problem: $\displaystyle \frac{x - 8}{x^2 - 4} + \frac{3}{x^2-2x}$

Answer: $\displaystyle \frac{x-3}{x(x+2)}, x\not=2$

What I ended up with: $\displaystyle \frac{x^2-5x+6}{x(x^2 - 4)}$

Any help on how to do problems like this would be great as well. I feel like I was doing everything correctly, but now that I'm at the end I don't see any way to make my answer look like the one in the book.
• Jun 25th 2013, 11:29 AM
HallsofIvy
Very close. All you need to do is observe that $\displaystyle x^2- 5x+ 6= (x- 2)(x- 3)$.

Even if you are not good at factoring in general, you should recognize that $\displaystyle x^2- 4= (x- 2)(x+ 2)$, which you apparently did to get the "least common denominator" and then check for common factors in the numerator and denominator. And you can do that by evaluating $\displaystyle 0^2- 5(0)+ 6= 6$ so x- 0= x is not a factor. $\displaystyle (-2)^2- 5(-2)+ 3= 17$ so x+ 2 is not a factor. But $\displaystyle 2^2- 5(2)+6= 0$ so x- 2 is a factor of $\displaystyle x^2- 5x+ 6$.
• Jun 25th 2013, 07:39 PM
Prove It
Quote:

Originally Posted by fogownz
Problem: $\displaystyle \frac{x - 8}{x^2 - 4} + \frac{3}{x^2-2x}$

Answer: $\displaystyle \frac{x-3}{x(x+2)}, x\not=2$

What I ended up with: $\displaystyle \frac{x^2-5x+6}{x(x^2 - 4)}$

Any help on how to do problems like this would be great as well. I feel like I was doing everything correctly, but now that I'm at the end I don't see any way to make my answer look like the one in the book.

Notice that \displaystyle \displaystyle \begin{align*} x^2 - 4 = (x - 2)(x + 2) \end{align*} and \displaystyle \displaystyle \begin{align*} x^2 - 2x = x(x - 2) \end{align*}. So the lowest common denominator is \displaystyle \displaystyle \begin{align*} x(x-2)(x+2) \end{align*}. You should get into the habit of keeping things factorised until the end, as it makes cancelling a lot easier.