Dividing rational expressions | Precalculus | Khan Academy

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. If we were to divide the fraction 6 over 25 by the fraction 15 over 9, we know that we can rewrite this as 6 over 25 divided by 15 over 9, which we know is the same thing as 6 over 25 times 9 over 15.

Then we can factor the various numerators and the denominators. This is 2 times 3; 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. Actually, that's about as far as we can get.

So then we'll have two times three times 3, which is going to be 18 in the numerator, and then in the denominator, we have 5 times 5 times 5, 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. First, I can just rewrite this as x squared minus 3x minus 4 all of that over negative 3x minus 15, and that's getting divided by this business: x squared minus 16 over 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 3x minus 4. Let's see, negative 1; let's see, negative 4 times plus 1 would be negative 4, and then these would add up to negative 3.

So I can rewrite this as x minus 4 times x plus 1. Rewrite it that way, and then I can rewrite what I have down here. I could factor out a negative 3, so I could write that as negative 3 times x plus 5.

And then I could write this one here. This is a difference of squares. It's going to be x plus 4 times x minus 4. And then, last but not least, this one over here: let's see, if I have a 5 and a negative 6, negative 6 plus 5 is negative 1, negative 6 times 5 is negative 30, so there's going to be x minus 6 times x plus 5.

Now, before I go any further, 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 0, that would be undefined. So we want to constrain our domain in that way.

So we know, for example, that x cannot be equal to negative 5. That would make this denominator equal to 0. Let me write that here: so x cannot be equal to negative 5. We also know that x cannot be equal to 6. x cannot be equal to 6, and this would also tell us that x cannot be equal to negative 5.

So I don't have to rewrite that again, but we're not done. 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, because you can't divide by zero.

So anything that would make this numerator equal to zero, which was this numerator right over here, would also make us 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 4, and actually, x cannot be equal to positive 4. 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 4 times x plus 1 all of that over negative 3 times x plus 5. And now I'm just going to, instead of divide by this, I'm going to multiply by the reciprocal.

So this is going to be times, and I'm just going to take the reciprocal: x minus 6 times x plus 5 all of that over x plus 4 times x minus 4. And once again, our domain is constrained in this way, but we see we have an x minus 4 in the numerator, now x minus 4 in the denominator, x plus 5 in the denominator, x plus 5 in the numerator.

And now we can say that this is going to be equal to x plus 1 times x minus 6 all of that over negative 3 times x plus 4. So the way it's written now, it's clear that x cannot be equal to negative 4, so this information you can say it's already there in this expression.

Now that we've reduced it to lowest terms, but this other information right over here 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 5, 6, or positive 4." You could throw the negative 4 in there if you like, but that one's already in the expression, so to speak.