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## Synthetic division of polynomials

# Why synthetic division works

CCSS.Math:

## Video transcript

What I want to do
now is simplify the exact same
expression but do it with traditional
algebraic long division. And hopefully we'll see why
synthetic division actually gives us the exact same result. We'll be able to
see the connections between synthetic division
and algebraic long division. So let's get started. So if we're doing
algebraic long division and set it up right over
here, the first thing we want to think about
is how many times does the highest degree
term here, which is our x, go into the highest
degree term here, which is our 3x to
the third power. Well x goes into 3x to the
third power 3x squared times. So we'll write it in
the x squared place, 3x squared times. Now you might already
see your parallel. When we did the
synthetic division, we dropped this 3
straight down, and this 3 represented 3x squared. So this 3 and this
3x squared are really representing the same thing. But you might be
saying, well here we have to do some thinking. We had to say x goes into 3x
to the third 3x squared times. Here, we just mindlessly
dropped this 3 straight down. How did that work? The reason why we were able to
mindlessly drop this 3 straight down is because
we assumed, to do this basic type of
synthetic division, that we had just an
x right over here. We didn't have a 3x. We didn't have a 4x. We didn't have an x squared. We just had an x. And if you divide an x into
whatever this highest degree term is, your
coefficient is going to be the exact same thing. It's just going to
be one degree lower. So you go from a 3x
to the third to 3x squared-- so the exact
same coefficient. And now, in our little synthetic
division right over here, that right over there
is the x squared term. So you went from 3x to
the third to 3x squared. We essentially divided it by x. And we could blindly do it
because we knew, we assumed, that we were dealing
with just a 1x to the first right over here. But let's keep on going
and see the parallels and see why we're essentially
doing the exact same thing. Now let's take this 3x
squared and multiply that times x plus 4. So 3x squared times
x-- I'll do it in that white color-- it's
going to be 3x to the third. And 3x squared times 4 is
going to be 12x squared. And now we'll want
to subtract this. So now we subtract. We subtract this out. These guys cancel
out, and you have 4x squared minus 12x squared. And so that will give
you negative 8x squared. So once again, you're probably
seeing some parallels. You had the 4x
squared over here. You have the 4x
squared over there. We just wrote the coefficient,
but that's what it represented. 4x squared, we
wrote the 4 there. Then we essentially
subtracted 12x squared. And the way we got that 12,
we multiplied 3 times 4, and then we subtracted. Here we're multiplying
3 times negative 4. We're essentially multiplying
3 times 4 and then subtracting. That's why we put
that negative there, so we don't have
to keep remembering to subtract this row. So we could just
keep adding them. But that's essentially
what we did. We multiplied 3 times this 4. And now we subtract it. We get that negative
12x squared. And then we subtracted, and
you got negative 8x squared. And you might say, oh, is
this the same negative 8 as this right over here? Not quite yet,
because over here, this negative 8 literally
represents negative 8x. This is actually part
of our simplification. When we divide
this into that, we got 3x squared minus 8x plus 30. So over here in the
algebraic long division, we then say, how many
times does x plus 4 go into negative 8x squared? Well x goes into negative 8x
squared negative 8x times. So this is the key
right over here. x goes into it negative 8x times. And once again, the
reason why we could just put this negative
8 here is we know that we are dividing
just by a 1x. So you're going to have
the exact same coefficient, just one degree lower. So this right over
here is our x term, and you see it right over
there, just like that. So a lot of the simplification
just comes from the idea that we are assuming with
the synthetic division that this is a 1x. But let's just keep going. So you have a negative 8x times
this business right over here gives you negative 8x times
x is negative 8x squared. And then you have negative 8x
times 4, which is negative 32x. And we can bring down all
of this business right over here, just so it
becomes a little bit simpler. So you have a negative 2x. And then over here,
you have a minus 1. And once again,
when you're doing traditional algebraic
long division, you're going to subtract
this from that up there. So if we're going
to subtract, that's like adding the negative. These characters cancel out. You have a negative 2x plus 32x. That gives us a positive 30x. And then we can bring down
that negative 1 if we want. I'll do that yellow
color, actually. We'll bring down
that negative 1. So this 30 has the
same coefficient here. But this 30 should be up here. This is going to be part
of our final answer. And to get that, once again,
it all comes from the fact that we know that
we had an x here when we did the
synthetic division. 30x divided by x is
just going to be 30. That 30 and this 30 is
the exact same thing. And then we multiply. 30 times x is 30x. Actually, let me write the
30x in that white color because that's the convention
I've been using-- 30x. 30 times 4 is 120. And then we are going to
subtract this from that. So we get negative 1
minus 120 is negative 121, which, right over there,
is our remainder-- which is exactly what
we got over there. So hopefully you
see the connection. Because we are assuming
that we are dividing by x plus or minus
something, we were able to make some
simplifying assumptions. Whenever you divide
this by an x, you know it's going to have
the same coefficient, just one degree lower. And we kept doing that. And so it allowed us to do
it a little bit simpler, a little bit faster,
and using less space.