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CCSS.Math: ,

so up here we are multiplying two rational expressions and here we're dividing one rational expression by another one and what I encourage you to do is pause these videos think about what these become when you multiply them out it may be you simplify it a little bit and I also want you to think about what constraints do you have to put on the 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 what I'm talking about so this is going to be in our numerator we are going to get 6x to the third power times two in our denominator we're going to have five times 3x 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 that both the numerator and denominator divisible by three so divide 6 by 3 you get to divide 3 by 3 you get 1 and we are left with 2x squared times 2 which is going to be 4 x squared over 5 times 1 times 1 over 5 and we could always write that as 4/5 x squared now if someone just presented you on the street with the expression 4/5 x squared and say for what X is is this defined Y well I could put any X here X could be 0 because 0 squared is 0 times 4/5 is just going to be 0 so it doesn't seems to be defined for 0 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 0 then you would be dividing by 0 right over here which would make this undefined so you can explicitly call it out X cannot be equal to 0 and so if you want this one to be algebraically equivalent you would have to make that same condition X cannot be equal to 0 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 6 X to the third over 5 times 2 over times 2 over 3x 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 they're going to get 2 divided by 0 its undefined but if you said okay well 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 you 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 x is anything other than zero and it's not defined for zero and so you'd say f of 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 this division situation here so immediately when you look at this you say well what it constraints here well X cannot be equal to zero because if X was zero this second this 5x to the fourth over four would be zero and you'd be dividing you'd be dividing by 0 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 2x to the fourth power over seven times the reciprocal times the reciprocal of this is going to be four over 5x to the fourth which is going to be equal to in the numerator we're going to have 8 X to the fourth so we're going to have eight X to the fourth 4 times 2x to the fourth over seven times five 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 in the denominator are divisible by X to the fourth so let's divide by X to the fourth and we get 8 over 35 so once again you just look at eight-thirty 15 well this is 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 B 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 30 fifths well now G of 0 would be defined as 8 30 fifths which would make it a different function so to make them algebraically equivalent you could say G of X is equal to eight thirty fifths 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 have simplified it