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## Synthetic division of polynomials

# Intro to polynomial synthetic division

CCSS.Math:

## Video transcript

In this expression, we're
dividing this third degree polynomial by this
first degree polynomial. And we could simplify
this by using traditional algebraic
long division. But what we're going
to cover in this video is a slightly
different technique, and we call it
synthetic division. And synthetic division
is going to seem like a little bit of voodoo
in the context of this video. In the next few
videos we're going to think about why it actually
makes sense, why you actually get the same result as
traditional algebraic long division. My personal tastes are not
to like synthetic division because it is very,
very, very algorithmic. I prefer to do traditional
algebraic long division. But I think you'll see that
this has some advantages. It can be faster. And it also uses a lot
less space on your paper. So let's actually perform
this synthetic division. Let's actually simplify
this expression. Before we start, there's
two important things to keep in mind. We're doing, kind of,
the most basic form of synthetic division. And to do this most basic
algorithm, this most basic process, you have
to look for two things in this
bottom expression. The first is that it has to
be a polynomial of degree 1. So you have just an x here. You don't have an x
squared, an x to the third, an x to the fourth or
anything like that. The other thing is, is that
the coefficient here is a 1. There are ways to do it if
the coefficient was different, but then our synthetic
division, we'll have to add a little bit of
bells and whistles to it. So in general, what
I'm going to show you now will work if
you have something of the form x plus or
minus something else. So with that said,
let's actually perform the synthetic division. So the first thing
I'm going to do is write all the coefficients
for this polynomial that's in the numerator. So let's write all of them. So we have a 3. We have a 4, that's
a positive 4. We have a negative 2. And a negative 1. And you'll see different people
draw different types of signs here depending on how they're
doing synthetic division. But this is the
most traditional. And you want to leave
some space right here for another row of numbers. So that's why I've gone
all the way down here. And then we look
at the denominator. And in particular, we're
going to look whatever x plus or minus is,
right over here. So we'll look at, right over
here, we have a positive 4. Instead of writing a positive 4,
we write the negative of that. So we write the negative,
which would be negative 4. And now we are
all set up, and we are ready to perform
our synthetic division. And it's going to
seem like voodoo. In future videos, we'll
explain why this works. So first, this first
coefficient, we literally just bring it straight down. And so you put the 3 there. Then you multiply what you
have here times the negative 4. So you multiply it
times the negative 4. 3 times negative
4 is negative 12. Then you add the 4
to the negative 12. 4 plus negative
12 is negative 8. Then you multiply negative
8 times the negative 4. I think you see the pattern. Negative 8 times negative
4 is positive 32. Now we add negative
2 plus positive 32. That gives us positive 30. Then you multiply the positive
30 times the negative 4. And that gives you negative 120. And then you add the negative
1 plus the negative 120. And you end up with
a negative 121. Now the last thing
you do is say, well, I have one term here. And in this plain,
vanilla, simple version of synthetic division, we're
only dealing, actually, when you have x plus
or minus something. So you're only going
to have one term there. So you separate out one term
from the right, just like that. And we essentially
have our answer, even though it
seems like voodoo. So to simplify this, you get,
and you could have a drum roll right over here,
this right over here, it's going to be
a constant term. You could think of it
as a degree 0 term. This is going to be an x term. And this is going to
be an x squared term. You can kind of just
build up from here, saying this first one is
going to be a constant. Then this is going to be an
x term, then an x squared. If we had more you'd have
an x to the third, an x to the fourth, so
on and so forth. So this is going to be equal
to 3x squared minus 8x plus 30. And this right over
here you can view as the remainder, so minus
121 over the x plus 4. This didn't divide perfectly. So over the x plus 4. Another way you could have
done it, you could have said, this is the remainder. So I'm going to have a
negative 121 over x plus 4. And this is going to be plus
30 minus 8x plus 3x squared. So hopefully that
makes some sense. I'll do another example
in the next video. And then we'll think about
why this actually works.