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### Course: Get ready for AP® Calculus>Unit 1

Lesson 22: Multiplying and dividing rational expressions

# Dividing rational expressions

When we divide rational expressions, we multiply the dividend (the first expression) by the reciprocal of the divisor (the second expression). We can also see if we can reduce the quotient to lowest terms. This is very similar to dividing fractions, only we also have to think about the domain while we do it. Created by Sal Khan.

## Want to join the conversation?

• I think there is a mistake here. If you graph this expression on Desmos, the x=6 is defined. So I'm guessing the expression should be seen directly as a multiplication hence the yellow expression is not in the denominator. And Yes, the others, x=-4,x=4,x=-5 are undefined when you graph it on Desmos.
• You are correct, Sal does not consider that the denominator of the denominator (when dividing fractions) actually becomes the numerator of the simplification. Thus, it is not really in the denominator, but numerator. Since this video appears to be new and you are the first to discover this, you should go to the help center and report this error so they can add a correction box.
• How would you be able to tell which numbers to exclude
• At the mark, after the expression is factored, Sal list the constraints for the numerator in purple that make the expression zero. However, he says one doesn't have to list any constraints that are in green because he says you can divide zero by other things. I don't understand!

(1 vote)
• We are dealing with a fraction that is being divided by a second fraction.

If the denominator of either fraction is zero, then that fraction (and thereby the entire expression) is undefined.
So, neither the expression in red nor the expression in yellow can be equal to zero.

If the numerator of the second fraction is zero and the denominator isn't, then that fraction will evaluate to zero,
which means we are dividing the first fraction by zero
and the entire expression is again undefined.
So, the expression in purple can't be equal to zero either.

This leaves us with the possibility that the expression in green is zero, while none of the others are.
This way, the first fraction will evaluate to zero and the second fraction will evaluate to some non-zero number,
which means we are dividing zero by a non-zero number
and thereby the entire expression will evaluate to zero.
So, the green expression can be equal to zero.
(1 vote)
• It's not wrong if I take the reciprocal from the get go right? I end up with the same answer, but wondering if its applicable to all problems or should I just do it last
(1 vote)
• if the the others, x=-4,x=4,x=-5 are undefined when you graph it on Desmos. i don't undersatnd
(1 vote)
• I don't think I understand why the 6 value is undefined. Someone else mentioned that there is no point on desmos that is undefined at 6. When I did the math work myself before the video, I did not get a 6, but maybe I skipped something? when doing keep change flip I skipped a middle step and just went to the final result (pre-crossing out the cancelable variants)
(1 vote)
• The original expression is undefined at 𝑥 = 6 due to division by zero.

𝑓(𝑥) = [(𝑥² − 3𝑥 − 4)∕(−3𝑥 − 15)]∕[(𝑥² − 16)∕(𝑥² − 𝑥 − 30)]
⇒ 𝑓(6) = (14∕−33)∕(20∕0) = (14∕−33)∕undefined = undefined.

– – –

Desmos is quite unconventional in the regard that it to some extent allows division by infinity, even though infinity isn't a defined number.

I have no idea why the developers made this decision, since all it really does is cause confusion.
(1 vote)
• Is there a video using the "(a/b)/(c/d) = ad/bc" method?
(1 vote)
• Dividing x and y is the same as multiplying x by the reciprocal of y (1/y). So (a/b)/(c/d)=(a/b)*(d/c)=(a*d)/(b*c)
(1 vote)

## Video transcript

- [Instructor] The goal of this video is to take this big, hairy expression, where we are essentially dividing rational expressions and see if we can essentially do the division and then write it in reduced terms. So if you are so inspired, I encourage you to pause the video and work on this on your own before we do this together. All right, now let's do this together. So this is completely analogous to dividing fractions. So if we were to divide the fraction six over 25 by the fraction 15 over nine, we know that we can rewrite this as six over 25 divided by 15 over nine, which we know is the same thing as six over 25 times nine over 15. And then we can factor the various numerators and the denominators. This is two times three. This is three times three. This is five times five. This is five times three. Let's see, three in the numerator, three in the denominator. And actually, that's about as far as we can get. So then we'll have two times three times three, which is going to be 18 in the numerator. And then in the denominator, we have five times five times five, which is 125. So we're going to do the exact same thing up here, but there's one extra complication. We have to keep track of the x values that would make this expression undefined in any way, because as we reduce to lowest terms, we might lose that information, but if we lose that information, then we have changed the expression. So we have to keep track of how we are constraining this domain. So first, I can just rewrite this as x squared minus three x minus four, all of that over negative three x minus 15. And that's getting divided by, divided by this business. X squared minus 16 over, we have x squared minus x minus 30. Now, the next thing we can do is we can factor the various numerators and denominators and think about which x values could get us into trouble. So x squared minus three x minus four. Let's see, negative one. Let's see, negative four times plus one would be negative four, and then these would add up to negative three. So I can rewrite this as x minus four times x plus one. Rewrite it that way. And then I can rewrite what I have down here, I could factor out a negative three. So I could write that as negative three times x plus five. And then I could write this one here. This is a difference of squares. It's going to be x plus four times x minus four. And then last but not least, this one over here. Let's see, if I have a five and a six, negative six plus five is negative one, negative six times five is negative 30. So it's going to be x minus six times x plus five. Now, before I go any further, and the reason why I factored at that point, is now we can think about what x values will get us in trouble. We know that any x values that make any of the denominators equal to zero, that would be undefined. So we wanna constrain our domain in that way. So we know, for example, that x cannot be equal to negative five. That would make this denominator equal to zero. Let me write that here. So x cannot be equal to negative five. We also know that x cannot be equal to six. X cannot be equal to six. And this would also tell us that x cannot be equal to negative five. So I don't have to rewrite that again. But we're not done. So we've figured out the x values that make these denominators equal to zero, but remember, we're also dividing by this entire expression here. So anything that would make the entire expression equal to zero would also be a problem, 'cause you can't divide by zero. So anything that would make this numerator equal to zero, which was this numerator right over here, would make us also divide by zero. So we have to constrain there. Not this numerator here, that one's fine. That one could be equal to zero. You can divide zero by other things. So let's see, we could see that x cannot be equal to negative four, and actually, x cannot be equal to positive four. So now we've fully constrained our domain and now we can proceed. So let me box this off right over here, and then we continue. I can rewrite all of this. So we're going to have x minus four times x plus one, all of that over negative three times x plus five, and now I'm just going to, instead of divide by this, I'm gonna multiply by the reciprocal. So this is going to be times, and I'm just gonna take the reciprocal, x minus six times x plus five, all of that over, we have x plus four times x minus four. And once again, our domain is constrained in this way, but we see we have an x minus four in the numerator now, x minus four in the denominator, x plus five in the denominator, x plus five in the numerator. And now we can say that this is going to be equal to x plus one times x minus six, all of that over negative three, negative three, times x plus four, x plus four. So the way it's written now, we would still, it's clear that x cannot be equal to negative four. So this information, you can say that it's already there in this expression now that we have reduced it to lowest terms, but this other information right over here, this has been lost. So if you want this expression to truly be equivalent to this expression up here, you would also have to say comma x cannot be equal to negative five, six, or positive four. You could throw the negative four in there if you like, but that one's already in the expression, so to speak.