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# Dividing quadratics by linear expressions (no remainders)

CCSS.Math:

## Video transcript

let's say someone walks up to you the on the street and they give you this expression x squared plus 7x plus 10 divided by X plus 2 and they say see if you could simplify this thing and so pause this video and see if you can do that and one way to think about it is what is x squared plus x squared plus 7x plus 10 divided by X plus 2 what is that going to be all right now there's two ways that you could approach this one way is to try to factor the numerator and see if it has a factor that is common to the denominator so let's try to do that so we've done this many many times if this is if this looks new to you I encourage you to review factoring polynomials other places on Khan Academy but what two numbers add up to 7 and when you multiply them you get 10 well that would be 2 and 5 so we could rewrite that numerator is X plus 2 times X plus 5 and then of course the denominator you still have X plus 2 and then we clearly see we have a common factor and so as long as X does not equal negative 2 because if x equals negative 2 this whole expression is undefined because then you get a 0 in the denominator so as long as X does not equal negative 2 well then we can divide the numerator and the denominator by X plus 2 once again the reason why I put that constraint is we can't divide the numerator in the denominator by 0 so for any other values of x this X plus 2 will be nonzero and we could divide the numerator and the denominator by that and they would cancel out and we would just be left with X plus 5 so another way to think about it is this expression our original expression could be viewed as X plus 5 for any X that is not equal to negative 2 now the other way that we could approach this is through algebraic long division which is very analogous to the type of long division that you might remember from I believe it was fourth grade so what you do is you say alright I'm gonna divide X plus 2 into x squared plus 7x plus 10 and in this technique you look at the highest degree terms so that you have an X there and an x squared there and say how many times does X go into x squared well it goes X times now you'd write that in this column because X is just X to the first power you could view this as the first degree column it's analogous to the place values that we talked about when we first learn numbers or how we regroup or about place value but here you could view it as degree places or something like that and then you take that X and you multiply it times this entire expression so x times 2 is 2x put that in the first degree column x times X is x squared and then what we want to do is we want to subtract these things in yellow from what we originally had in blue so we could do it this way and then we will be left with 7x minus 2x is 5x and then x squared minus x squared is just a zero and then we can bring down this +10 and once again we look at the highest degree term X goes into 5x 5 times that's a zero degree it's a constant so I'll write it in the constant column 5 times 2 is 10 5 times X is 5 and then I'll subtract these from what we have up here and notice we have no remainder what's interesting about algebraic long division we'll probably see in another video too you can actually have a remainder so those are going to be situations where just the factoring technique alone would not have worked in this situation this this model would have been easier but this is another way to think about it you say look X plus 2 times X plus 5 is going to be equal to this now if you wanted to rewrite this expression the way we did here and say hey this expression is equal to X plus 5 we would have to constrain the domain you'd say hey for all X is not equal to negative 2 for these to be completely identical expressions