# Thread:

1. ## Dividing Rational Expressions

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.

2. 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?

3. The answer for problem 10 is 5 but I don't see how that equals 0.

Problem 11, factor the top and bottom?

Problem 13, factor again?

4. 10. 5, -1/2

11. -(x+3)/2

12. (x+3)/(x-3)

5. Would it be to much trouble to ask how you solved those?

6. Originally Posted by juvenilepunk
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.

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] $x-5=0$

[2] $2x+1=0$

Now solve these two equations and you will find the two values of x that makes the original rational expression undefined.

7. Originally Posted by masters
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] $x-5=0$

[2] $2x+1=0$

Now solve these two equations and you will find the two values of x that makes the original rational expression undefined.

Thank you for the explanation, got it. Thanks.

8. Originally Posted by juvenilepunk
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.

Now, for [12], you have to do a little factoring in the numerator and denominator.

$\frac{x^3-9x}{6x-2x^2}=\frac{x(x^2-9)}{2x(3-x)}$

Notice that now in the numerator you have the difference of two squares.
Remember how to factor this type?

$\frac{x(x-3)(x+3)}{2x(3-x)}$

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: $\frac{x(x-3)(x+3)}{-2x(x-3)}$

I'll bet you can finish this up now.

Try to tackle [13], and let us know where you get stuck.
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