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Current time:0:00Total duration:5:53

CCSS.Math:

let's see if we can simplify this expression so pause the video and have a try at it and then we're going to do it together right now all right so when you look at this it looks like both the numerator in the denominator they might you might be able to factor them and maybe they have some common factors that you can divide the numerator and the denominator by to simplify it so let's first try to factor the numerator X to the fourth plus 8x squared plus 7 and at first it might be a little intimidating because you have an X to the fourth year it's not a it's not a quadratic it's a fourth degree polynomial but like any if you like a lot of quadratics that we've seen in the past it does seem to have a pattern for example if this said x squared plus plus 8x plus 7 you'd say oh well this is pretty straightforward to factor what two numbers add up to 8 and when I take the product I get 7 well there's only two numbers that were you taking the product and you get positive 7 that are going to be positive and and and and they need to be positive if they're going to add up to positive 8 and that's 1 in 7 so this would be X plus 7 times X plus 1 well if you just think of instead of thinking in terms of X and x squared if you just think in terms of x squared and X to the fourth it's going to be the exact same thing so this thing can be written as x squared plus 7 times x squared plus 1 if you want you could do some type of a substitution saying saying that a is equal to x squared in which case so if you said that a is equal to x squared then this thing would become a squared plus 8 a plus 7 and then you would factor this into a plus 7 and a plus 1 and then you would you would undo the substitution and that's x squared plus 7 x squared plus 1 but hopefully you see what's going on here this this is this is the the higher order term and then this is half the degree of that so it fits this mold and so you could do a substitution or you could just recognize well okay instead of I'm dealing dealing with X Squared's I'm dealing with X to the fourth all right so that's the numerator now let's think about let's think about the denominator so the denominator both of these terms are divisible by three X so let's let's factor out a 3x so it's 3x times 3x times if you factor out a 3x here 3 divided by 3 is 1 X to the fifth divided by X is X to the 4th and then if you subtract er out a 3x you're just going to get 1 and so far this doesn't seem too helpful I don't see an X to the 4th minus 1 or 3 X in the numerator but maybe I can factor this out further X to the 4th minus 1 and that's because it is a difference of squares and you might say wait wait I'm always used to recognizing a difference of squares it's something like a squared minus 1 which you could write as a plus 1 times a minus 1 well this would be a squared minus 1 if you say that if you say that a is equal to x squared then this would be a squared minus 1 so let's rewrite all of this so let's rewrite so this is all going to be equal to same numerator so you must do it in green same numerator x squared plus 7 can't factor that out anymore times x squared plus 1 can't factor that out anymore all that over 3x but this I can views the difference of squares so this is x squared squared and this is obviously 1 squared so this is going to be x squared plus 1 times x squared minus x x squared minus 1 now cleared I have an x squared in the numerator x squared minus 1 in the numerator x squared sorry x squared plus 1 in the numerator x squared plus 1 in the denominator and so I could cancel them out and I'm going to be left with in the numerator x squared plus 7 over 3x times x squared minus 1 now this looks pretty simple and we want to be a little careful because whenever we do this cancelling out we don't want to make too make sure that we restrict the the X's for which the expression is defined if we want them to be algebraically equivalent so this one would this be this would obviously be undefined if so X cannot be equal to 0 X cannot be equal to plus or minus one pot positive or negative one would make this expression right over here equals zero so it cannot be equal to zero X cannot be equal to I'll write plus or minus one that would make this part zero but this right over here this one unless we're assuming we will deal only with real numbers this one can't ever equal zero if you're dealing with real numbers because x squared is always going to be non-negative and you're adding it to a positive value and so this part this this factor would have never made the entire thing undefined so we can actually just factor it out or cancel it out without worrying much about it and so this is actually algebraically equivalent to what we had originally now we could write these constraints on it if we want if someone were to ask me you know for what X is this expression not defined well it's clear it's not defined for X that would make the denominator equal zero dividing by zero not defined or if X is plus or minus one it would make the denominator equal zero but that is that comes straight out of this expression so this expression and our original expression are algebraically equivalent now if you want to do you could expand the bottom out a little bit you can multiply it out if you like that's equivalent to so x squared plus seven over 3x times x squared is 3x to the third minus 3x so these are all these are all equivalent expressions and and we are done