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# Multiplying & dividing rational expressions: monomials

CCSS.Math:

## Video transcript

- [Voiceover] So up here, we are multiplying two rational expressions. And here, we're dividing one rational expression by another one. Now what I encourage you do is pause this video and think about what these become when you multiply them. I don't know, maybe you simplify it a little bit, and I also want you to think about what constraints do you have to put on the x values in order for your resulting expression to be algebraically equivalent to your original expression. So let's work it out together just so you realize what I'm talking about. So this is going to be, in our numerator, we are going to get six x to the third power times two, and our denominator, we're going to have five times three x. And 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 to the third by x. We get x squared. And we can also see the both the numerator and denominator are divisible by three, so divide six by three, you get two. Divide three by three, you get one. And we are left with two x squared times two, which is going to be four x squared over five times one times one over five. And we can also write that as 4/5 x squared. Now someone just presented you on the street with the expression 4/5 x squared and say, for what x is this defined? I could put any x here, x could be zero because zero squared is zero times 4/5 is just going to be zero, so it does seems to be defined for zero, and that is true. But if someone says, how would I have to constrain this in order for it to be algebraically equivalent to this first expression? Well then, you'd have to say, well, 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 can not be equal to zero. And so if you want this 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, if you said f of x is equal to six x to the third over five times two over, times two over three x; and if someone said, well what is f of zero, you would say f of zero is undefined. Undefined. Why is that? Because you put x equals zero there, you're going to get two divided by zero and it's undefined. But if you said, okay, well, can I simplify this a little bit to get the exact same function? Well, we're saying you can say f of x is equal to 4/5 times x squared. But if you just left it at that, you would get f of zero is equal to zero. 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 them, to make it clear that this is equivalent to that one, you would have to say x cannot be equal to zero. Now these functions are equivalent because now, if u said f of zero, you'd say all right, x cannot be equal to zero, you know? This would be the case if x is anything other than zero and it's not defined for zero, and so you would say f or zero is undefined. So 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, woah, what are constrains here? Well, x cannot be equal to zero because if x was a zero, this second, this five x to the fourth over four would be zero and you'd be dividing, you'd be dividing by zero. So we can explicitly call out that x cannot be equal to zero. And so if x cannot be equal to zero in the original expression, if the result, 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. So this is going to be the same thing as two x to the fourth power over seven times the reciprocal, times... The reciprocal of this is going to be four over five x to the fourth, which is going to be equal to in the numerator, we're going to have eight x to the fourth. So we're going to have eight x to the fourth, four times two x to the fourth, over seven times five x to the fourth is 35 x to the fourth. And now, there's something. We can do a little bit of simplification here, both the numerator and the denominator are divisible by x to the fourth, so let's divide by x to the fourth and we get eight over 35. So once again, you just look at eight 30, 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 does not, cannot be equal to zero. And to see, you know, 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 of x is equal to, is equal to all of this business here. Well, g of zero would be undefined. But if you said g of x is equal to 8/35, well now, g of zero would be defined as 8/35, which would make it a different function. So to make them algebraically equivalent, you could say g of x is equal to 8/35, as long as x does not equal zero. And you could say it's undefined if you want. Undefined for x equals zero. Or you don't 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 had simplified it.