Dividing quadratics by linear expressions with remainders: missing x-term
About this video
This video divides x^2 + 1 by x + 2, where the numerator has no x term. First it rewrites x^2 + 1 as (x^2 - 4) + 5 = (x + 2)(x - 2) + 5 using a difference of squares, giving x - 2 + 5/(x + 2). Then it uses algebraic long division, writing a 0x placeholder, and gets the same answer: x - 2 with a remainder of 5.
You'll be able to
- rewrite a numerator using a difference of squares so the divisor becomes a factor.
- use a zero placeholder term (like 0x) to keep degree columns lined up in long division.
- divide x^2 + 1 by x + 2 and write the answer as x - 2 + 5/(x + 2).
Key words
- difference of squares
- An expression like x^2 - 4 that factors as (x + 2)(x - 2).
- placeholder term
- A term with a zero coefficient, like 0x, written to keep each degree in its own column.
- remainder
- What is left after dividing as far as possible; here it is 5.
Transcript
- [Instructor] 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 one divided by x plus two." 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 one, it's not obvious that you can factor it out. But can you write something that has x plus two as a factor, and interestingly enough has no first degree terms? 'Cause 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 different of squares using x plus two. So we know that x plus two times x minus two is equal to x squared minus four. So what is we were to write x squared minus four 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 four as x plus two times x minus two. And so then I could rewrite this entire expression as x plus two times x minus two, all of that over x plus two plus five, plus five over x plus two.
And now as long as x does not equal negative two, then we could divide the numerator and the denominator by x plus two. And then we would be left with x minus two plus five over x plus two, and I'll put that little constraint, if I wanna say that this expression is the same as that first expression, for x does not equal, for x not equaling negative two.
And so here, we'd say "Hey! X squared plus one divided by x plus two is x minus two," and then we have a remainder of five, remainder of five. 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 two into x squared plus one. Now when I write things out I like to be very careful with my, I guess you could say, with my different places for the different degrees. So x squared plus one has no first degree term, so I'm gonna write the one out here.
So second degree, no first degree term, and then we have a one, which is a zero degree term, or constant term. And so, we do the same drill, how many times does x go into x squared. We're just looking 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 two is two x. X times x is x squared. And now we wanna subtract. And so what is this gonna be equal to? We know the x squared's cancel out. And then I'm gonna be subtracting negative two x from, you could do this as plus zero x up here plus one, and so you're left with negative two x.
And then we bring down that one plus one. X goes into negative two x, negative two times. Put that in the constant column. Negative two times two is negative four. And then negative two times x is negative two x. Now we have to be very careful here because we want to subtract the negative two x minus four from the negative two x plus one.
We could view it as this or we could just distribute the negative sign. And then this will be positive two x plus four. And then, the two x's, the two x and the negative two x cancels out. One plus four is five 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 two, x minus two with a remainder of five.
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