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CCSS.Math: ,

- So we have a polynomial here. What I'm curious about
is what is the remainder if I were to divide this polynomial by, let's just say, x minus,
I want the remainder when I divide this
polynomial by x minus two? You could do this. You could figure this out with algebraic long division, but I'll give you a hint. It is much simpler and
much less computation intensive and takes much less space on your paper if you use the polynomial remainder theorem. If that's unfamiliar to
you, there's other videos that actually cover that. So why don't you have a go at it. All right, so now let's
work through this together. The polynomial remainder theorem tells us that when I take a polynomial, p of x, and if I were to divide
it by an x minus a, the remainder of that is just going to be equal to p of a. Is just going to be equal to p of a. So in this case, our p of x is this. What is our a? Well our a is going to be positive two. Remember it's x minus a. So let me do this. Our a is equal to positive two. So to figure out the remainder, we just have to evaluate p of two. So let's do that. So the remainder in this case is going to be equal to p of two, which is equal to, so let's see, it's going to be, I'll
just do it all in magenta, negative three times
eight minus, let's see, minus four times four plus 20 minus seven. So let's see, this is -24 minus 16 plus 20 minus seven. So that gives us, let's see, -24 minus 16, this is -40. All right, I'm just doing it step by step. This is equal to negative,
actually I can do this in my head. All right, here we go. So this is -40 plus 20 is
-20 minus seven is -27. That was pretty neat
because if we attempted to do this without the
polynomial remainder theorem, we would have had to do a bunch of algebraic long division. Now if we did the algebraic long division, we would have gotten the
quotient and all of that, but we don't need the quotient, we don't need to know. So if we did all the
algebraic long division, you know, we would have taken our p of x and then we would have
divided the x minus a into it, and we would have
gotten a quotient here, q of x, and we would have done all this business down here, all this
algebraic long division. Probably wouldn't have
even fit on the page. But eventually we would
have gotten to a point where we got an expression that has a lower degree than this. It would have to be a
constant because this is a 1st degree, so it
would have be essentially a zero degree. So we would have eventually
gotten to our -27. But this was much, much, much, much easier then having to go through
this entire exercise. Hopefully you appreciated that.