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Dividing polynomials with remainders

Sal divides (x^3+5x-4) by (x^2-x+1) using long division. Created by Sal Khan and Monterey Institute for Technology and Education.

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Video transcript

We divide x to the third plus 5x minus 4 by x minus x-- actually, I think this is supposed to be an x squared. Let me correct it. So we're going to divide x to the third plus 5x minus 4 by x squared minus x plus 1. And just to see the different ways we can rewrite this. We could rewrite this as x to the third plus 5x minus 4 divided by x squared minus x plus 1. Or maybe the best way to write it in this circumstance, since we're going to do algebraic long division, is to write it as x squared minus x plus 1, divided into x to the third plus-- and actually, I'm going to leave some blank space here. We don't have an x squared term here. But I'm going to leave some space for it, just so that we can align everything in the proper place when we actually do the division. So x to the third plus nothing to the x squared power plus 5x minus 4. So we have a place for the third power, the second power, the first power, and the 0-th power. So now let's just do a little bit of algebraic long division. Let's look at the highest-degree term. x squared goes into x to the third how many times? Well, it goes into it x times. x to the third divided by x squared is equal to x to the 3 minus 2, which is equal to x to the 1, which is equal to x. So it goes x times. I'll write the x right over here. And we multiply x times this entire thing. x times x squared is x to the third. x times negative x is negative x squared. x times 1 is positive x. And now we want to subtract this whole expression from that whole expression. And that's the same thing as adding the opposite, or multiplying each of these terms by negative 1, and then adding it to these terms. So let's do that. So we have negative x to the third. Negative 1 times negative x squared is positive x squared. And then positive x times negative 1 is negative x. And so let's now add everything. x to the third minus x to third-- those cancel out. 0 plus x squared gives us an x squared. 5x minus x gives us a plus 4x. And then we bring down this minus 4. We're not adding anything to it there. You could view there's a 0 here. So let's bring down the minus 4. And now let's look at the highest-degree terms. x squared goes into x squared exactly one time. It's the same thing, so we put a plus 1. And then you have 1 times x squared is x squared. 1 times negative x is negative x. 1 times 1 is 1. And now we want to subtract this from that, or we want to add the opposite. And to add the opposite, we can just multiply each of these terms by negative 1. x squared becomes negative x squared. Negative x times negative 1 is positive x. And then positive 1 times negative 1 is negative 1. Now let's do the addition. x squared minus x squared-- they cancel out. 4x plus x is 5x. And then we have negative 4 minus 1 is negative 5. Now, you might be tempted to keep dividing, but you can't any more. This term right here, the highest-degree term here, is now higher than the highest-degree term that you're going to try to divide into. So we have a remainder. So the answer to this is-- this expression right over here is equal to x plus 1 plus the remainder, plus 5x minus 5-- whatever the remainder is-- divided by x squared minus x plus 1. If this was divisible, we could keep dividing, but we're saying it's not. It's now a lower degree than this down here. So we could say it's x plus 1 plus whatever this remainder is, divided by this thing over here. So our answer-- I'm going to write it one more time. It's x plus 1 plus 5x minus 5, over x squared minus x plus 1. And we can check that this works. If we take this thing over here, and we multiply it by this thing over here, we should get the x to the third plus 5x minus 4. So let's do that. Let's multiply this thing by x squared minus x plus 1. And to do that, let's just distribute this whole trinomial times each of these terms. When we do the first term, we have x squared minus x plus 1 times x. So that's going to be x times x squared, which is x to the third; x times negative x, which is negative x squared; x times 1, which is plus x. Then we can multiply this whole thing times 1. So it's going to be plus x squared minus x plus 1. I'm just multiplying all of these times 1. And then we can multiply this whole thing times this thing. Now, this is the same as the denominator here. So it'll cancel out. This will cancel with that. And we're just going to be left with the numerator over here, so plus 5x minus 5. And now we can try to simplify it. We only have one third-degree term, the x to the third. So we have x to the third here. Second-degree terms-- we have a negative x squared. And then we also have a positive x squared. So they cancel out with each other. First-degree terms-- let's see. We have a positive x and a negative x. Those cancel out with each other. So we're just going to have that 5x over here. So we're just going to have this 5x. So then we have plus 5x. And then we have the 0-th degree terms, or the constant terms. We have a positive 1 and a negative 5. Add them together. You get negative 4. So you get x to the third plus 5x minus 4, which is exactly what we had over here.