Dividing quadratics by linear expressions with remainders: missing x-term | Algebra 2 | Khan Academy

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.

All right, so there's two ways to approach this. Either we can try to re-express the numerator where it involves an x plus two 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 write something that has x plus 2 as a factor? Interestingly enough, it has no first degree terms because we don't want some first degree, weird first degree terms sitting up there. 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 five to get to plus one? So what if we were to write x squared minus four, and then we write plus five? 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. So then I could rewrite this entire expression as ( \frac{x + 2 \cdot (x - 2)}{x + 2} + \frac{5}{x + 2} ). And now, as long as x does not equal negative 2, then we could divide the numerator and the denominator by x plus 2.

Then we would be left with ( x - 2 + \frac{5}{x + 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 not equaling negative 2. So here we'd say, "Hey, x squared plus 1 divided by x plus 2 is ( x - 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 one. 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 one has no first degree term, so I'm going to write the one out here—so, second degree, no first degree term, and then we have a one, which is you could view as our zero degree term or our constant term.

And so we do the same drill. How many times does x go into x squared? We just look at the highest degree terms here. x goes into x squared x times; that's first degree, so I put it in the first degree column. x times 2 is 2x, x times x is x squared.

And now we want to subtract. So what is this going to be equal to? We know the x squareds can 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.

Plus 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 would be positive 2x plus 4.

And then the two x's, the 2x and the negative 2x cancels out; 1 plus 4 is 5. And there's no obvious way of dividing x plus 2 into 5, so we would call that the remainder. Exactly what we had before when we divided with algebraic long division; we got ( x - 2 ) with a remainder of five.