Once again I have no one to turn to for help so I must put my fate in your hands.
These problems are for a study guide for my midterm tonight and I have no idea how to do them. Any help is appreciated, thank you.
For Problem 10, you should find the values of x which make the denominator zero. Can you see why that would be the thing to do?
For problem 11, what do you suppose would be a good first step?
For problem 13, there might be an easier way than multiplying everything out. What do you suppose that might be?
Hi juvenilepunk,
In order for the expression to be undefined, the denominator would have to equal zero.
Now how can that happen? Two ways:
[1] $\displaystyle x-5=0$
[2] $\displaystyle 2x+1=0$
Now solve these two equations and you will find the two values of x that makes the original rational expression undefined.
Now, for [12], you have to do a little factoring in the numerator and denominator.
$\displaystyle \frac{x^3-9x}{6x-2x^2}=\frac{x(x^2-9)}{2x(3-x)}$
Notice that now in the numerator you have the difference of two squares.
Remember how to factor this type?
$\displaystyle \frac{x(x-3)(x+3)}{2x(3-x)}$
Now, it gets a little tricky. Notice the one of the factors in the denominator is almost like one of the factors in the numerator, but not quite.
The signs are reversed (3 - x) instead of (x - 3).
It would be nice if we could do something to remedy that. And we can. We just factor out a -1 and it switches. (3 - x) = -(x - 3).
Now we have: $\displaystyle \frac{x(x-3)(x+3)}{-2x(x-3)}$
I'll bet you can finish this up now.
Try to tackle [13], and let us know where you get stuck.