Long Division of Polynomials Example 1
Uploaded by TheIntegralCALC on 15.06.2011
Transcript:
Long Division of Polynomials Example 1
Hi, everyone!
Welcome back to integralcalc.com.
Today we're going to do a long division problem with polynomials.
So we've been given this fraction, three x cubed minus two x squared plus four x minus three
divided by another polynomial, x squared plus three x plus three.
And we need to perform long division with this polynomials in order to simplify the fraction.
So this is no different than the long division that you would have learned in elementary school
except that this time, we're going to be dividing polynomials.
So the first thing you know is that you always... remember with long division, we've got our formula like this.
We’re going to put the numerator inside so we've got three x cubed minus two x squared plus four x minus three.
And then we're going to put the denominator on the outside, so x squared plus three x plus three, right?
And just like we've learned with long division before,
you start this problem by figuring out what to multiply by x squared inorder to get three x cubed.
So we would have to multiply x squared by three x to get three x cubed, right?
And the reason we do that is because now that we figured out that it's three x,
we're going to take three x and multiply it through this part here.
So we'll do three x times each of these terms, so three x times x squared gives us three x cubed.
Then we'll do three x times three x which gives us a positive nine x squared, and then three x times three gives us a positive nine x.
So once we've multiplied through, we're going to subtract this polynomial from the one above it.
So when we do that, you notice that we get three x cubed minus three x cubed
so we get a zero out in front and that was the whole point of making sure that this three x here was something
that we could multiply by x squared to match three x cubed here.
Because we wanted these to cancel. So we'll get zero.
We’ll get negative two x squared.
This will be a minus nine x squared because of the negative sign out in front here.
So negative two x squared minus nine x squared is a negative eleven x squared
and then this will be a minus nine x because we've got four x minus...
out in front here... minus nine x.
So that will give us a negative five x.
And then once we've performed that subtraction, we always want to bring down the next term in our original numerator,
so this three, negative three comes down and we end up with negative three. Like that.
And now we perform the same concept again.
So what will we have to multiply by x squared to get 11 x squared?
Of course, that would be negative eleven.
So once we've identified that, we go ahead and multiply negative eleven through each term of x squared plus three x plus three.
So we get negative eleven x squared minus thirty-three x, and then minus thirty-three.
And once we've done that, we go ahead and put this in parenthesis and we're going to be subtracting.
Always. You always subtract.
So obviously, we get negative eleven x squared because we have the double negative here, negative and negative,
we'll get negative eleven x squared plus eleven x squared which will be zero.
So notice, we have the canceling again.
Then we get a negative five x instead of a minus thirty-three x, it'll be a double negative again
so we'll get negative five x plus thirty-three x and that'll give us a positive twenty-eight x.
Then negative three minus a negative thirty-three would be negative three plus thirty-three which would give us a positive thirty.
So now at this point, notice we don't have any more terms out here to bring down. Right?
So we're done with that part.
We could figure out what to multiply by x squared to get twenty-eight x but if we did that,
we realize that we have to multiply by twenty-eight divided by x.
Twenty-eight divided by x is what you would have to multiply by x squared to get twenty-eight x.
And whenever we run into that, whenever we realize that we need to multiply by a fraction,
we know that we've gotten to the end where we're going to call the rest of what's left here, the remainder.
Because this part, multiplying by a fraction like that starts getting too complicated so what we do is we stop there
and we say that the simplified polynomial is now first this part; whatever we got up here,
everything up here so that's three x minus eleven.
So we'd say that it's three x minus eleven plus the remainder which is what's left down here.
So twenty-eight x plus thirty, the remainder divided by everything here that was outside our division bracket.
So we get x squared plus three x plus three, right?
Our original denominator.
So this is the part from the top here and then this whole thing right here, the second part is the remainder.
So we can call that our final answer.
That’s as far as we get in terms of the long division.
I hope that video helped you, guys and I'll see you in the next one.
Bye!
Hi, everyone!
Welcome back to integralcalc.com.
Today we're going to do a long division problem with polynomials.
So we've been given this fraction, three x cubed minus two x squared plus four x minus three
divided by another polynomial, x squared plus three x plus three.
And we need to perform long division with this polynomials in order to simplify the fraction.
So this is no different than the long division that you would have learned in elementary school
except that this time, we're going to be dividing polynomials.
So the first thing you know is that you always... remember with long division, we've got our formula like this.
We’re going to put the numerator inside so we've got three x cubed minus two x squared plus four x minus three.
And then we're going to put the denominator on the outside, so x squared plus three x plus three, right?
And just like we've learned with long division before,
you start this problem by figuring out what to multiply by x squared inorder to get three x cubed.
So we would have to multiply x squared by three x to get three x cubed, right?
And the reason we do that is because now that we figured out that it's three x,
we're going to take three x and multiply it through this part here.
So we'll do three x times each of these terms, so three x times x squared gives us three x cubed.
Then we'll do three x times three x which gives us a positive nine x squared, and then three x times three gives us a positive nine x.
So once we've multiplied through, we're going to subtract this polynomial from the one above it.
So when we do that, you notice that we get three x cubed minus three x cubed
so we get a zero out in front and that was the whole point of making sure that this three x here was something
that we could multiply by x squared to match three x cubed here.
Because we wanted these to cancel. So we'll get zero.
We’ll get negative two x squared.
This will be a minus nine x squared because of the negative sign out in front here.
So negative two x squared minus nine x squared is a negative eleven x squared
and then this will be a minus nine x because we've got four x minus...
out in front here... minus nine x.
So that will give us a negative five x.
And then once we've performed that subtraction, we always want to bring down the next term in our original numerator,
so this three, negative three comes down and we end up with negative three. Like that.
And now we perform the same concept again.
So what will we have to multiply by x squared to get 11 x squared?
Of course, that would be negative eleven.
So once we've identified that, we go ahead and multiply negative eleven through each term of x squared plus three x plus three.
So we get negative eleven x squared minus thirty-three x, and then minus thirty-three.
And once we've done that, we go ahead and put this in parenthesis and we're going to be subtracting.
Always. You always subtract.
So obviously, we get negative eleven x squared because we have the double negative here, negative and negative,
we'll get negative eleven x squared plus eleven x squared which will be zero.
So notice, we have the canceling again.
Then we get a negative five x instead of a minus thirty-three x, it'll be a double negative again
so we'll get negative five x plus thirty-three x and that'll give us a positive twenty-eight x.
Then negative three minus a negative thirty-three would be negative three plus thirty-three which would give us a positive thirty.
So now at this point, notice we don't have any more terms out here to bring down. Right?
So we're done with that part.
We could figure out what to multiply by x squared to get twenty-eight x but if we did that,
we realize that we have to multiply by twenty-eight divided by x.
Twenty-eight divided by x is what you would have to multiply by x squared to get twenty-eight x.
And whenever we run into that, whenever we realize that we need to multiply by a fraction,
we know that we've gotten to the end where we're going to call the rest of what's left here, the remainder.
Because this part, multiplying by a fraction like that starts getting too complicated so what we do is we stop there
and we say that the simplified polynomial is now first this part; whatever we got up here,
everything up here so that's three x minus eleven.
So we'd say that it's three x minus eleven plus the remainder which is what's left down here.
So twenty-eight x plus thirty, the remainder divided by everything here that was outside our division bracket.
So we get x squared plus three x plus three, right?
Our original denominator.
So this is the part from the top here and then this whole thing right here, the second part is the remainder.
So we can call that our final answer.
That’s as far as we get in terms of the long division.
I hope that video helped you, guys and I'll see you in the next one.
Bye!