Dividing rational expressions: unknown expression | High School Math | Khan Academy

We're told the following equation is true for all real values of Y for which the expression on the left is defined, and D is a polynomial expression. They have this equation here; what is D?

All right, so essentially what they're saying is they don't want us to somehow solve this equation. They're saying D is going to be some type of a polynomial expression. They tell us that right over there: D is a polynomial expression. If we figure out what D is, this left-hand side of the expression is going to evaluate to one for all real values of Y for which the expression is defined.

So let's think about how we would tackle that. Well, the first thing that pops into my mind, if I'm dividing by a fraction or a rational expression, that's the same thing as multiplying by the reciprocal. So let's just rewrite this on the left-hand side. This is (20y^2 - 80 / D) times... Oh, let me do the reciprocal. Let me be careful! Times the reciprocal of this. If I divide by something, it's the same thing as multiplying by the reciprocal. So let me just swap the numerator and denominator.

All right, (y^3 + 9y^2) all of that over (4y^2 - 8y). That's going to be equal to 1. Now, let's see if we can simplify all of this business on the left-hand side a little bit. So let's see; over here I can divide both terms by 20. Let me factor out 20 because I think then it's going to end up being a difference of squares.

If I factor out a 20, this is going to be the same thing as (20(y^2 - 4)), and (y^2 - 4) we can rewrite as ((y + 2)(y - 2)). It is a difference of squares, so let me write that: ((y + 2)(y - 2)).

All right, this down here (4y^2 - 8y), well, it looks like we can factor out a 4Y. So this is going to be the same thing as (4y(y - 2)). All right, so let me cross that out, so it's the same thing as (4y(y - 2)), and I already see this (y - 2) here and this (y - 2) here are going to cancel out.

And let's see; up here both terms are divisible by (y^2). So I can rewrite this as... I don't know if this is actually going to be helpful because it's going to... If you factor... Well, let me just do it just in case. So that's the same thing as (y^2(y^2)(y + 9)). All right?

And so we can rewrite all of these things. If we were to multiply everything together, we would end up getting in the numerator (20(y + 2)(y - 2)(y^2)(y + 9)). I'm just multiplying all the numerators. And that's going to be over, in the denominator I would have whatever the expression D is, times (4y(y - 2)).

And that's all going to be equal to 1. Now let's think about it: we can divide; we have (y - 2 / y - 2), so those cancel out. Let's see; we can divide the numerator and the denominator by y, so that would just become 1, and then that would just become (y) to the first power.

And so what we’d be left with in the numerator is (20(y)(y + 2)(y + 9)) over (4D). This is equal to 1. Now if we want to solve for D, well, we could just multiply both sides by D, and (1 \times D) is just going to be D. So you're going to have (D = 5(y)(y + 2)(y + 9)), and we're done.

This is D: this is the polynomial—whoops, that is the polynomial expression that we are looking for. If you were to substitute this back in and then try to simplify it, well, you would end up with all of this over here, and D would be this, and so it would all just cancel out, and you would be left with one for all real values y for which the expression on the left is actually defined.

And, you know, there are some values of Y for which the expression on the left is not defined. If Y is equal to zero, this denominator is zero, and you're dividing by zero. Well, that's not defined. And then when you multiply by the reciprocal, if this were to become zero, that wouldn't be cool either.

There are multiple ways to make this equal to zero. Y could be equal to 9; that would also make this bottom zero. So we could think about that if we wanted to, but they're not asking us to do that. They're saying for all real values for y for which the expression is defined, find the D that makes all of this business equal to one, and we just did that.