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Multiplying rational expressions | Precalculus | Khan Academy


4m read
·Nov 10, 2024

So what I have here is an expression where I'm multiplying rational expressions, and we want to do this multiplication and then reduce to the lowest term. So if you feel so inspired, I encourage you to pause this video and see if you can have a go at that.

All right, now let's work through this together. So multiplying rational expressions like this, it's very analogous to multiplying fractions. For example, if I were to multiply six twenty-fifths times fifteen over nine, there's a few ways you could do it. You could just multiply six times fifteen in the numerator and twenty-five times nine in the denominator.

But the way that many of us approach it, so that it's easier to reduce to lowest terms, is to factor things. To realize that, look, six is two times three, nine is three times three, fifteen is three times five, and twenty-five is five times five. Then you can realize in your eventual product you're going to have a 5 in the numerator and a 5 in the denominator. You can see that 5 divided by 5 is 1, 3 divided by 3 is 1, and then 3 divided by 3 is 1.

So all you'd be left with is that 2 and then that 5. So this is going to be equal to two-fifths. We'll do the analogous thing here with these rational expressions. We're going to factor all of them in the numerators and the denominators, and then we'll see if we can divide the numerator and the denominator by the same thing.

Now, the one thing we have to make sure of as we do that is we keep track of the domain because these rational expressions here might have x values that make their denominators equal to zero. Even if we reduce to lowest terms and we get rid of those expressions, in order for the expressions to be the same expression, we have to constrain the domain in the same way.

So let's get started. So this is going to be equal to—I’ll just rewrite everything. x squared minus nine, how do we factor that? Well, that's going to be a difference of squares. We could write that as x plus 3 times x minus 3, and then that is going to be over this business.

Let's see, 5 squared is 25, negative 5 plus negative 5 is negative 10. So this is going to be x minus 5 times x minus 5. If what I'm just doing here with the factoring is not making sense, I encourage you to review factoring on Khan Academy.

Then we multiply that times, let's see, in this numerator here, I can factor out a 4. So that's going to be 4 times x minus 5, which is going to be useful. I have an x minus 5 there, x minus 5 there, and then that's going to be over. Let's see, this expression over here, 2 plus 3 is 5; 2 times 3 is 6. So it's going to be x plus 2 times x plus 3.

Now, before I start reducing to lowest terms, let's think about the domain here. The domain is going to be constrained by things that make these denominators equal to 0. So the domain would be all real numbers except x cannot equal five.

Let me write it over here: x cannot equal five. Because if that happened, then this denominator would be equal to zero. x cannot be equal to negative two; x could not be equal to negative 2 because that would make the denominator here zero, which would make the denominator here zero. And x cannot be equal to negative three.

So the domain is constrained in this way. We have to carry this throughout, no matter what we do to the expression. This is the constraints on our domain. With that out of the way, now we can reduce to lowest terms.

So x, we have an x plus 3 in the numerator, x plus 3 in the denominator, x minus 5 in the numerator, x minus 5 in the denominator, and I think we've gone about as far as we can. So when we multiply the numerators, we are going to get this business. It’s going to be 4 times x minus 3 over, we have an x minus 5 here, x minus 5, and then we have an x plus 2.

We have an x plus 2. And we could leave it like this if you want. In some cases, people like to multiply the things out, but we're done. We've just finished multiplying these rational expressions.

We have to remind ourselves that x cannot be equal to any of these things. Now, the way that we've simplified it, we still have an x minus 5 right over here, so it might be redundant to say that x cannot be equal to 5 because that's still the case in our reduced terms expression here.

And that's true also of the x cannot be equal to negative 2. We still have an x plus 2 here, so still even in this expression it's pretty clear that x cannot be equal to negative 2. However, that x cannot equal negative 3 isn't so obvious when you just look at this expression.

But in order for this expression to be completely equivalent to the original, it has to have the same domain. And so you might want to explicitly say that x cannot be equal to negative 3 here. You could also say the other two, but those are, that's still very clear when you look at this expression.

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