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Current time:0:00Total duration:4:32

Dividing polynomials by linear expressions: missing term

Video transcript

in front of us we have another screenshot from Khan Academy and I've modified a little bit so I have a little bit of extra space it says divide the polynomials the form of your answer should either be a straight-up polynomial or polynomial plus the remainder over X minus 5 which we have here in the denominator where P of X is a polynomial and K is an integer so we've done stuff like this but like always I encourage you to pause this video and work on this on your own and if you were doing this on Khan Academy there's a little bit of an input box here where you'd have to type in the answer but let's just do it on paper for now all right so we're trying to figure out what X minus 5 divided into 2 X to the third power and actually I want to be careful here because I want to be very very organized about my different degree columns so this is my third degree column and then I want my second degree column but there is no second degree term here there's a first degree term so I'll write it out here so minus 47 X and actually to be even more careful I'll write plus 0x squared and then I have minus 15 by putting that plus 0 x squared that's making sure I'm doing good I guess degree place column hygiene all right so now we can work through this and first we could say hey how many times does X go into the highest degree term here well X goes into 2 X to the third power 2x squared times and we'd want to put that in the second degree column 2x squared you can see how it would have gotten messy if I put negative 47 X Jeb like where do I put that 2x squared and you might have gotten you might confuse yourself which none of us would want to happen all right 2x squared times negative 5 is negative 10 x squared 2x squared times X is 2x to the third power now we want to subtract what we have in red from what we have in blue so I'll multiply them both by negative 1 so that becomes a negative and then that one becomes a positive and that's actually one of the biggest areas for careless errors if you have negative here and you just want to subtract it because you know you have to subtract we're like no I'm subtracting a negative it needs to be a positive now all right so 0 x squared plus 10x squared is 10x squared and then the 2 X to the 3rd minus 2x to the 3rd is just a zero and then we can bring down that negative 47 X and once again we look at the highest degree terms X goes into 10x squared 10x times so plus 10x 10x times negative 5 is negative 50 X negative 50 X 10x times x is 10x squared and once again we want to subtract what we have in teal from what we have in red so we can multiply both of these times negative 1 that becomes a negative this one becomes a positive now negative 47 X plus 50 X is positive 3x and then 10 x squared minus 10x squared gets cancelled out bring down that 15 come on down I used to watch a lot of prices right going growing up never quite made it to the show all right X goes into 3x how many times it goes 3 times 3 times negative 5 is negative 15 3 times X is 3x we want to subtract the orange from the teal and so this becomes a negative this becomes a positive 15 my or negative 15 plus 15 is a 0 and 3 X minus 3 X is 0 so you're just left with 0 so no remainder so this whole thing you could react I as 2 x squared plus 10 X plus 3 and once again if this was on Khan Academy there would be a little bit of an input box that looks something like this and you would have to type this in now if you wanted these to be exactly the same expression you would also need to constrain the domain you would say okay for X does not equal positive 5 and the reason why you have to constrain that is where the whole the whole reason why we can even divide by X minus 5 is we're assuming that X minus 5 is not equal to 0 and it's generally not equal to 0 as long as X does not equal to positive 5 but for the sake of that exercise you don't need to constrain the domain like this