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In this expression, we're
dividing this third degree

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polynomial by this
first degree polynomial.

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And we could simplify
this by using

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traditional algebraic
long division.

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But what we're going
to cover in this video

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is a slightly
different technique,

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and we call it
synthetic division.

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And synthetic division
is going to seem

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like a little bit of voodoo
in the context of this video.

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In the next few
videos we're going

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to think about why it actually
makes sense, why you actually

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get the same result as
traditional algebraic long

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division.

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My personal tastes are not
to like synthetic division

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because it is very,
very, very algorithmic.

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I prefer to do traditional
algebraic long division.

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But I think you'll see that
this has some advantages.

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It can be faster.

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And it also uses a lot
less space on your paper.

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So let's actually perform
this synthetic division.

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Let's actually simplify
this expression.

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Before we start, there's
two important things

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to keep in mind.

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We're doing, kind of,
the most basic form

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of synthetic division.

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And to do this most basic
algorithm, this most basic

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process, you have
to look for two

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things in this
bottom expression.

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The first is that it has to
be a polynomial of degree 1.

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So you have just an x here.

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You don't have an x
squared, an x to the third,

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an x to the fourth or
anything like that.

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The other thing is, is that
the coefficient here is a 1.

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There are ways to do it if
the coefficient was different,

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but then our synthetic
division, we'll

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have to add a little bit of
bells and whistles to it.

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So in general, what
I'm going to show

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you now will work if
you have something

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of the form x plus or
minus something else.

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So with that said,
let's actually

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perform the synthetic division.

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So the first thing
I'm going to do

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is write all the coefficients
for this polynomial

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that's in the numerator.

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So let's write all of them.

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So we have a 3.

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We have a 4, that's
a positive 4.

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We have a negative 2.

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And a negative 1.

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And you'll see different people
draw different types of signs

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here depending on how they're
doing synthetic division.

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But this is the
most traditional.

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And you want to leave
some space right here

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for another row of numbers.

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So that's why I've gone
all the way down here.

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And then we look
at the denominator.

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And in particular, we're
going to look whatever

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x plus or minus is,
right over here.

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So we'll look at, right over
here, we have a positive 4.

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Instead of writing a positive 4,
we write the negative of that.

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So we write the negative,
which would be negative 4.

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And now we are
all set up, and we

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are ready to perform
our synthetic division.

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And it's going to
seem like voodoo.

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In future videos, we'll
explain why this works.

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So first, this first
coefficient, we literally just

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bring it straight down.

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And so you put the 3 there.

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Then you multiply what you
have here times the negative 4.

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So you multiply it
times the negative 4.

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3 times negative
4 is negative 12.

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Then you add the 4
to the negative 12.

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4 plus negative
12 is negative 8.

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Then you multiply negative
8 times the negative 4.

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I think you see the pattern.

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Negative 8 times negative
4 is positive 32.

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Now we add negative
2 plus positive 32.

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That gives us positive 30.

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Then you multiply the positive
30 times the negative 4.

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And that gives you negative 120.

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And then you add the negative
1 plus the negative 120.

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And you end up with
a negative 121.

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Now the last thing
you do is say, well, I

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have one term here.

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And in this plain,
vanilla, simple version

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of synthetic division, we're
only dealing, actually,

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when you have x plus
or minus something.

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So you're only going
to have one term there.

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So you separate out one term
from the right, just like that.

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And we essentially
have our answer,

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even though it
seems like voodoo.

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So to simplify this, you get,
and you could have a drum roll

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right over here,
this right over here,

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it's going to be
a constant term.

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You could think of it
as a degree 0 term.

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This is going to be an x term.

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And this is going to
be an x squared term.

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You can kind of just
build up from here,

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saying this first one is
going to be a constant.

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Then this is going to be an
x term, then an x squared.

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If we had more you'd have
an x to the third, an x

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to the fourth, so
on and so forth.

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So this is going to be equal
to 3x squared minus 8x plus 30.

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And this right over
here you can view

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as the remainder, so minus
121 over the x plus 4.

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This didn't divide perfectly.

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So over the x plus 4.

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Another way you could have
done it, you could have said,

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this is the remainder.

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So I'm going to have a
negative 121 over x plus 4.

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And this is going to be plus
30 minus 8x plus 3x squared.

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So hopefully that
makes some sense.

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I'll do another example
in the next video.

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And then we'll think about
why this actually works.