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Current time:0:00Total duration:4:43

Dividing quadratics by linear expressions with remainders: missing x-term

Video transcript

this polynomial division business is a little bit more fun than we expected so let's keep going so let's say that I guess again someone walks up to you in the street and says what is x squared plus 1 divided by X plus 2 so pause this video and have a go at that and I'll give you a little bit of a warning this one's a little bit more involved than you might expect alright so there's two ways to approach this either we can try to re Express the numerator where it involves an X plus 2 somehow or we could try to do algebraic long division so let me do the first way so x squared plus 1 it's not obvious that you can factor it out but can you can you write something that has X plus 2 as a factor and interestingly enough has no first degree terms because we don't want some first degree weird first degree terms sitting up there and the best thing that I could think of is constructing a difference of squares using X plus 2 so we know that X plus 2 times X minus 2 is equal to x squared minus 4 so what if we were to write x squared minus 4 up here and then we would just have to add 5 to get 2 plus 1 so what if we were to write x squared minus 4 and then we write plus 5 this expression and that expression up there those are completely equivalent but why did I do that well now I can write x squared minus 4 as X plus 2 times X minus 2 and so then I could rewrite this entire expression as X plus 2 times X minus 2 all of that over X plus 2 plus 5 plus 5 over X plus 2 and now as long as X does not equal negative 2 that we could divide the numerator and the denominator by X plus 2 and then we would be left with X minus 2 plus 5 over X plus 2 and I'll put that little constraint if I want to say that this expression is the same as that first expression for X does not equal for X not equally negative 2 and so here would say hey x squared plus 1 divided by X plus 2 is X minus 2 and then we have a remainder of 5 remainder of 5 now let's do the same question or try to rewrite this using algebraic long division we'll see that this is actually a little bit more straightforward so we are going to divide X plus 2 into x squared plus 1 now when I write things out I like to be very careful with my I guess you could say my different places for the different degrees so x squared plus 1 has no first degree term so I'm gonna write the 1 out here so second degree no first degree term and then we have a 1 which is you could viewers are 0 degree term or our constant term and so we do the same drill how many times does X go into x squared we're just look at the highest degree terms here X goes into x squared x times that's first degree so I put in the first degree column x times 2 is 2x x times X is x squared and now we want to subtract and so what is this going to be equal to we know the X Squared's can't cancel out and then I'm going to be subtracting negative 2x from you could view this as plus 0x up here plus 1 and so you're left with negative 2x and then we bring down that 1+1 X goes into negative 2x negative 2 times put that in the constant column negative 2 times 2 is negative 4 and then negative 2 times X is negative 2x now we have to be very careful here because we want to subtract the negative 2x minus 4 from the negative 2x plus 1 we could view it as this or we could just distribute the negative sign and then this will be positive 2x plus 4 and then the 2x is the 2x and the negative 2x cancels out 1 plus 4 is 5 and there's no obvious way of dividing X plus two into five so we would call that the remainder exactly what we had before when we divided with algebraic long division we got X minus 2 X minus 2 with a remainder of 5