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多項式的除法 (英)

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多項式的除法 (英) : 多項式的除法

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In this video, we're going to learn to divide polynomials,
and sometimes this is called algebraic long division.
But you'll see what I'm talking about
when we do a few examples.
Let's say I just want to divide 2x plus 4 and
divide it by 2.
We're not really changing the value.
We're just changing how we're going to express the value.
So we already know how to simplify this.
We've done this in the past. We could divide the numerator
and the denominator by 2, and then this
would be equal to what?
This would be equal to x plus 2-- let me write it this way--
it would be equal to, if you divide this by 2,
it becomes an x.
You divide the 4 by 2, it becomes a 2.
If you divide the 2 by 2, you get a 1.
So this is equal to x plus 2, which is pretty
straightforward, I think.
The other way is you could have factored a 2 out of here,
and then those would have canceled out.
But I'll also show you how to do it using algebraic long
division, which is a bit of overkill for this problem.
But I just want to show you that it's not fundamentally
anything new.
It's just a different way of doing things, but it's useful
for more complicated problems.
So you could have also written this as 2 goes into 2x plus 4
how many times?
And you would perform this the same way you would do
traditional long division.
You'd say 2-- you always start with the highest degree term.
2 goes into the highest degree term.
You would ignore the 4.
2 goes into 2x how many times?
Well, it goes into 2x x times and you put
the x in the x place.
x times 2 is 2x.
And just like traditional long division, you now subtract.
So 2x plus 4 minus 2x is what?
It's 4, right?
And then 2 goes into 4 how many times?
It goes into it two times, a positive two times.
Put that in the constants place.
2 times 2 is 4.
You subtract, remainder 0.
So this might seem overkill for what was probably a
problem that you already knew how to do and
do it in a few steps.
We're now going to see that this is a very
generalizable process.
You can do this really for any degree polynomial dividing
into any other degree polynomial.
Let me show you what I'm talking about.
So let's say we wanted to divide x plus 1 into x squared
plus 3x plus 6.
So what do we do here?
So you look at the highest degree term here, which is an
x, and you look at the highest degree term here,
which is an x squared.
So you can ignore everything else.
And that really simplifies the process.
You say x goes into x squared how many times?
Well, x squared divided by x is just x, right? x goes into
x squared x times.
You put it in the x place.
This is the x place right here or the x to
the first power place.
So x times x plus 1 is what?
x times x is x squared.
x times 1 is x, so it's x squared plus x.
And just like we did over here, we now subtract.
And what do we get?
x squared plus 3x plus 6 minus x squared-- let me be very
careful-- this is minus x squared plus x.
I want to make sure that negative sign only-- it
applies to this whole thing.
So x squared minus x squared, those cancel out.
3x, this is going to be a minus x.
Let me put that sign there.
So this is minus x squared minus x, just to be clear.
We're subtracting the whole thing.
3x minus x is 2x.
And then you bring down the 6, or 6 minus 0 is nothing.
So 2x plus 6.
Now, you look at the highest degree term, an x and a 2x.
How many times does x go into 2x?
It goes into it two times.
2 times x is 2x.
2 times 1 is 2.
So we get 2 times x plus 1 is 2x plus 2.
But we're going to want to subtract this from this up
here, so we're going to subtract it.
Instead of writing 2x plus 2, we could just write negative
2x minus 2 and then add them.
These guys cancel out.
6 minus 2 is 4.
And how many times does x go into 4?
We could just say that's zero times, or we could say that 4
is the remainder.
So if we wanted to rewrite x squared plus 3x plus 6 over x
plus 1-- notice, this is the same thing as x squared plus
3x plus 6 divided by x plus 1, this thing divided by this, we
can now say that this is equal to x plus 2.
it is equal to x plus 2 plus the remainder divided by x
plus 1 plus 4 over x plus 1.
This right here and this right here are equivalent.
And if you wanted to check that, if you wanted to go from
this back to that, what you could do is multiply this by x
plus 1 over x plus 1 and it add the two.
So this is the same thing as x plus 2.
And I'm just going to multiply that times x plus
1 over x plus 1.
That's just multiplying it by 1.
And then to that, add 4 over x plus 1.
I did that so I have the same common denominator.
And when you perform this addition right here, when you
multiply these two binomials and then add the 4 up here,
you should you get x squared plus 3x plus 6.
Let's do another one of these.
They're kind of fun.
So let's say that we have-- we want to simplify x squared
plus 5x plus 4 over x plus 4.
So once again, we can do our algebraic long division.
We can divide x plus 4 into x squared plus 5x plus 4.
And once again, same exact process.
Look at the highest degree terms in both of them. x goes
into x squared how many times?
It goes into it x times.
Put it in the x place.
This is our x place right here.
X times x is x squared.
x times 4 is 4x.
And then, of course, we're going to want to subtract
these from there.
So let me just put a negative sign there.
And then these cancel out.
5x minus 4x is x.
4 minus 0 is plus 4.
x plus 4, and then you could even see this coming.
You could say x plus 4 goes into x plus 4 obviously one
time, or if you were not looking at the constant terms,
you would completely just say, well, x goes
into x how many times?
Well, one time.
Plus 1.
1 times x is x.
1 times 4 is 4.
We're going to subtract them from up here, so it cancels
out, so we have no remainder.
So this right here simplifies to-- this is
equal to x plus 1.
And there's other ways you could have done this.
We could have tried to factor this numerator.
x squared plus 5x plus 4 over x plus 4.
This is the same thing as what?
We could have factored this numerator as x plus 4
times x plus 1.
4 times 1 is 4.
4 plus 1 is 5, all of that over x plus 4.
That cancels out and you're left just with x plus 1.
Either way would have worked, but the algebraic long
division will always work, even if you can't cancel out
factors like that, even if you did have a remainder.
In this situation, you didn't.
So this was equal to x plus 1.
Let's do another one of these just to make sure that you
really-- because this is actually a very, very useful
skill to have in your toolkit.
So let's say we have x squared-- let me
just change it up.
Let's say we had 2x squared-- I could really make these
numbers up on the fly.
2x squared minus 20x plus 12 divided by-- actually, let's
make it really interesting, just to show you that it'll
always work.
I want to go above quadratic.
So let's say we have 3x to the third minus 2x squared plus 7x
minus 4, and we want to divide that by x squared plus 1.
I just made this up.
But we can just do the algebraic long division to
figure out what this is going to be or what this
simplified will be.
x squared plus 1 divided into this thing up here, 3x to the
third minus 2x squared plus 7x minus 4.
Once again, look at the highest degree term.
x squared goes into 3x to the third how many times?
Well, it's going to go into it 3x times.
You multiply 3x times this, you get 3x to the third.
So it's going to go into it 3x times.
So you have to write the 3x over here in the x terms. So
it's going to go into it 3x times, just like that.
Now let's multiply.
3x times x squared is 3x to the third, right?
3x times squared plus 3x times 1.
So we have a 3x over here.
I'm making sure to put it in the x place.
And we're going to want to subtract them.
And what do we have?
What do we have when we do that?
These cancel out.
We have a minus 2x squared.
And then 7x minus-- because I just subtracted 0 from there--
7x minus 3x is plus 4x, and we have a minus 4.
Once again, look at the highest degree term.
x squared and a negative 2x squared.
So x squared goes into negative 2x
squared negative 2 times.
Negative 2, put it in the constants place.
Negative 2 times x squared is negative 2x squared.
Negative 2 times 1 is negative 2.
Now, we're going to want to subtract these from there, so
let's multiply them by negative 1, or
those become a positive.
These two guys cancel out.
4x minus 0 is-- let me switch colors-- 4x minus 0 is 4x.
Negative 4 minus negative 2 or negative 4 plus 2 is equal to
negative 2.
And then x squared, now it has a higher degree than 4x, and
the highest degree here, so we view this as the remainder.
So this expression we could rewrite it as being equal to
3x minus 2-- that's the 3x minus 2-- plus our remainder
4x minus 2, all of that over x squared plus 1.
Hopefully, you found that as fun as I did.

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多項式的除法 (英)

多項式的除法 (英) : 多項式的除法