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CCSS.Math: ,

- [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.