Multiplying & dividing rational expressions: monomials | High School Math | Khan Academy

So up here we are multiplying two rational expressions, and here we're dividing one rational expression by another one. What I encourage you to do is pause these videos and think about what these become when you multiply them out. Maybe you simplify it a little bit. I also want you to think about what constraints you have to put on the X values in order for your resulting expression to be algebraically equivalent to your original expression.

Let's work it out together just so you realize what I'm talking about. In our numerator, we are going to get (6x^3 * 2). In our denominator, we're going to have (5 * 3x). We can see both the numerator and the denominator are divisible by (x), so let's divide the denominator by (x); we get one there. Let's divide (x^3) by (x); we get (x^2). We can also see that both the numerator and denominator are divisible by three. So divide (6) by (3); you get (2), and divide (3) by (3); you get (1).

We are left with (2x^2 * 2), which is going to be (\frac{4x^2}{5}) or ( \frac{4}{5} x^2). Now someone just presented you on the street with the expression (\frac{45}{x^s}) and said, "For what (X) is this defined?" You would say, "Well, I could put any (X) here. (X) could be zero. (0 * 45) is just going to be (0), so it doesn't seem to be defined for zero." That is true, but if someone asks how you would have to constrain this in order for it to be algebraically equivalent to this first expression, well then you would have to say, "This first expression is not defined for all (X). For example, if (x) were equal to zero, then you would be dividing by zero right over here, which would make this undefined."

So you could explicitly call it out: (X) cannot be equal to zero. If you want this one to be algebraically equivalent, you would have to make that same condition: (X) cannot be equal to zero. Another way to think about it: if you had a function defined this way, if you said (f(x) = \frac{6x^3}{5 * 2} \cdot \frac{2}{3x}), and someone said, "Well, what is (f(0))?" You would say (f(0)) is undefined. Why is that? Because you put (x = 0) there, you're going to get ( \frac{2}{0}); it's undefined.

But if you said, "Okay, well can I simplify this a little bit to get the exact same function?" Well, you can say (f(x) = \frac{45}{x^2}), but if you just left it at that, you would get (f(0) = 0), so now it would be defined at zero. But then this would make it a different function. These are two different functions the way they're written right over here. Instead, to make it clear that this is equivalent to that one, you would have to say (x) cannot be equal to (0).

Now these functions are equivalent, and now if you said (f(0)), you'd say, "Alright, (X) cannot be equal to (0); you know this would be the case if (x) is anything other than zero, and it's not defined for zero." So you would say (f(0)) is undefined. Now these two functions are equivalent, or these two expressions are algebraically equivalent.

So thinking about that, let's tackle this division situation here. So immediately, when you look at this, you say, "Well, what is the constraint here?" Well, (X) cannot be equal to zero. If (x) was zero, this second part, ( \frac{5x^4}{4}), would be zero, and you'd be dividing by zero. So we can explicitly call out that (X) cannot be equal to zero.

If (x) cannot be equal to zero in the original expression, whatever we get for the resulting expression in order for it to be algebraically equivalent, we have to give this same constraint. So let's multiply this, or let's do the division. This is going to be the same thing as (\frac{2x^4}{7}) times the reciprocal. The reciprocal of this is going to be (\frac{4}{5x^4}), which is going to be equal to, in the numerator, we're going to have (8x^4).

So we're going to have (\frac{8x^4}{7 * 5 * x^4}), which is (35x^4). Now there's something we can do, a little bit of simplification here. Both the numerator and the denominator are divisible by (x^4), so let's divide by (x^4), and we get (\frac{8}{35}).

So once again, you just look at (\frac{8}{35}). Well, this is going to be defined for any (x); (x) isn't even involved in the expression. But if we want this to be algebraically equivalent to this first expression, then we have to make the same constraint: (X) cannot be equal to (0).

Now, this even seems a little bit more nonsensical to say (x) cannot be equal to zero for an expression that does not even involve (X). But one way to think about it is, imagine a function that was defined as (g(X)) is equal to all of this business here. Well, (g(0)) would be undefined, but if you said (g(X) = \frac{8}{35}), well, now (g(0)) would be defined as (\frac{8}{35}), which would make it a different function.

So to make them algebraically equivalent, you could say (g(x) = \frac{8}{35}) as long as (X) does not equal zero. You could say it's undefined for (x = 0), or you'll even have to include that second row, and that will literally just make it undefined. But now this expression, this algebraic expression, is equivalent to our original one, even though we have simplified it.