WEBVTT mathematics/algebra-2/eaton
00:00:00.000 --> 00:00:02.000
Welcome to Educator.com.
00:00:02.000 --> 00:00:08.500
Today, we are going to be talking about dividing polynomials, starting out with a review of techniques learned in Algebra I,
00:00:08.500 --> 00:00:13.600
and then going on to learn a new technique called synthetic division.
00:00:13.600 --> 00:00:17.300
First, we are discussing dividing a polynomial by a monomial.
00:00:17.300 --> 00:00:22.800
The technique is to divide each term of the polynomial by the monomial.
00:00:22.800 --> 00:00:30.400
And if you think about just dividing with regular numbers, if you have something like this,
00:00:30.400 --> 00:00:41.300
you could handle it by saying, "15 divided by 3, plus 6 divided by 3, plus 24 divided by 3," and splitting that up.
00:00:41.300 --> 00:00:53.900
This would give you 5; 6 divided by 3 is 2; and then, 24 divided by 3 is 8; adding this up gives you 15.
00:00:53.900 --> 00:01:08.100
You may also have done this and added 15, 6, and 24; and those add up to 45/3, to get 15.
00:01:08.100 --> 00:01:18.500
So, these two are equivalent; and that is what allows you to handle dividing a polynomial by a monomial, by dividing each term by the monomial.
00:01:18.500 --> 00:01:28.400
For example, 10x⁴ + 8x³ - 12x, all divided by 2x:
00:01:28.400 --> 00:01:31.800
looking at this up here as my example of how to handle it, I am going to say,
00:01:31.800 --> 00:01:46.400
"OK, this is equivalent to 10x⁴ divided by 2x, plus 8x³ divided by 2x, minus 12x divided by 2x.
00:01:46.400 --> 00:01:50.700
And this is a step you might do in your head, or you might write it out.
00:01:50.700 --> 00:02:02.400
10 divided by 2 is 5; x⁴ divided by x would be the same as saying x^4 - 1, or x³.
00:02:02.400 --> 00:02:09.200
8 divided by 2 gives me 4; x³ divided by x is x².
00:02:09.200 --> 00:02:14.800
Here, I have -12 divided by 2 (is 6); and the x's cancel.
00:02:14.800 --> 00:02:20.600
My result is 5x³ + 4x² - 6.
00:02:20.600 --> 00:02:28.100
And I handled that by dividing each term in the polynomial by the monomial, separately.
00:02:28.100 --> 00:02:34.600
OK, now when you are working with dividing a polynomial by another polynomial, one technique is long division.
00:02:34.600 --> 00:02:41.500
And we talked about this in Algebra I, and you can also review those lectures; and I will review it here, as well.
00:02:41.500 --> 00:02:46.200
Just as with regular division with numbers (long division) you might end up with a remainder.
00:02:46.200 --> 00:02:53.000
So, first, just reviewing long division (which I am sure you know well, but just to think about the steps you are taking):
00:02:53.000 --> 00:02:55.600
you probably do this so much that you don't even think about the steps,
00:02:55.600 --> 00:03:00.300
but you want to realize what you are doing each step of the way,
00:03:00.300 --> 00:03:04.100
so that you can apply it when you are dividing polynomials.
00:03:04.100 --> 00:03:12.500
If you were asked to divide something like 513 by 2 by long division, think about what you would be doing.
00:03:12.500 --> 00:03:22.100
First, you divide with the first term, using the first term in the divisor: 5 divided by 2.
00:03:22.100 --> 00:03:24.400
OK, so that is going to give me 2.
00:03:24.400 --> 00:03:33.500
Next, I am going to multiply--I am going to multiply 2 by the divisor 2 to get 4.
00:03:33.500 --> 00:03:44.500
And then, I am going to subtract that product: 2 times 2 is 4, so I am going to subtract that--that is going to give me 1.
00:03:44.500 --> 00:03:55.200
Next, I will bring down the next number; in the case of polynomials, I will bring down the next term.
00:03:55.200 --> 00:04:14.800
OK, 2 goes into 11 five times; this gives me 5 times 2 (is 10): 11 minus 10 is 1; now I am bringing down this 3.
00:04:14.800 --> 00:04:25.100
2 goes into 13 six times, and this is going to give me 12, because now I am multiplying and then subtracting; and I have a remainder of 1.
00:04:25.100 --> 00:04:42.700
Remember that you can always check your answer by realizing that the dividend equals the divisor, times the quotient, plus the remainder.
00:04:42.700 --> 00:04:57.000
Looking at this with these numbers, my dividend here is 513; so it equals the divisor (which is 2), times 256, plus the remainder of 1, which equals 513.
00:04:57.000 --> 00:04:58.200
So, that checks out.
00:04:58.200 --> 00:05:01.600
Now, with polynomials, it's the same concept.
00:05:01.600 --> 00:05:16.300
Look at this example, which is going to be 5x² + 4x - 7, all divided by x + 3.
00:05:16.300 --> 00:05:24.400
Using long division: it is the same steps that I took over here.
00:05:24.400 --> 00:05:30.900
I am going to start out with x and divide that into 5x²; I have 5x² divided by x;
00:05:30.900 --> 00:05:37.400
and you can see that this is going to give me 5x, because one of the x's will cancel.
00:05:37.400 --> 00:05:50.000
OK, so that gives me 5x; now, my next step is to multiply 5x times x, which gives me 5x².
00:05:50.000 --> 00:06:02.300
5x times 3 is 15x; now, I divided; I multiplied; the next step is to subtract.
00:06:02.300 --> 00:06:09.700
The 5x's cancel out; 4x - 15x is -11x.
00:06:09.700 --> 00:06:18.100
The next step: bring down the next number--and here, that would be -7, the next term.
00:06:18.100 --> 00:06:23.300
OK, so I go back up; I was down here--I go back to #1 again and divide.
00:06:23.300 --> 00:06:31.200
-11x divided by x; the x's cancel, so that gives me -11.
00:06:31.200 --> 00:06:35.400
So, I am going to put that up here; I divided; now I need to multiply.
00:06:35.400 --> 00:06:48.800
-11 times x is -11x; -11 times 3 gives me -33, minus (I subtract--subtracting is the same as adding the opposite,
00:06:48.800 --> 00:06:57.100
so I am going to add, and the opposite of these two would be positive terms)...-11x and 11x cancels out.
00:06:57.100 --> 00:07:04.300
And then, I have -7 + 33, which is going to give me 26; and that is my remainder.
00:07:04.300 --> 00:07:09.400
So, my answer is the quotient, 5x - 11, plus the remainder of 26.
00:07:09.400 --> 00:07:26.000
Now, to check this out, I am going to say, "OK, the dividend equals the divisor, times the quotient, plus the remainder."
00:07:26.000 --> 00:07:41.800
Let's make sure this checks out: that is 5x² - 11x + 15x - 33 (just using the FOIL method) + 26.
00:07:41.800 --> 00:07:49.900
Simplify to get 5x²...-11x + 15x gives me 4x; -33 + 26 is -7.
00:07:49.900 --> 00:07:56.900
And this does check out; so again, this is really just using the same techniques as you used for long division with numbers.
00:07:56.900 --> 00:08:03.100
Now, one thing to be aware of is that sometimes there are missing terms.
00:08:03.100 --> 00:08:10.000
And if that is the case, you need to use a coefficient of 0 and represent those terms when you are dividing.
00:08:10.000 --> 00:08:20.200
Now, if you think about it, when we have numbers like, say, 107, we have a 0 here as a placeholder.
00:08:20.200 --> 00:08:25.600
When we are doing long division with polynomials, we need to do the same thing.
00:08:25.600 --> 00:08:37.600
For example, let's say you were asked to divide 3x³ + 6x - 4: you were asked to divide that by x + 2.
00:08:37.600 --> 00:08:43.700
If you look here, there is a missing term: I have an x³ term, but I have no x² term.
00:08:43.700 --> 00:08:45.500
And then, I have an x and a constant.
00:08:45.500 --> 00:08:57.200
So, I have a missing x² term; and really, I can represent that x² term by giving it a coefficient of 0.
00:08:57.200 --> 00:08:59.300
And I need to do that before I divide.
00:08:59.300 --> 00:09:05.700
So, if I have x + 2, that is my divisor; my dividend--what I am going to write here--is
00:09:05.700 --> 00:09:15.300
3x³ + (my missing term) 0x² + 6x - 4.
00:09:15.300 --> 00:09:18.800
And then, I am going to go about long division like I usually do.
00:09:18.800 --> 00:09:26.300
So, I take 3x³, and I divide that by x; and that is going to give me 3x²,
00:09:26.300 --> 00:09:33.700
because 3x³ divided by x is going to give me 3x².
00:09:33.700 --> 00:09:43.800
OK, so I divided; the next step is to multiply: 3x² times x is 3x³; 3x² times 2 is 6x².
00:09:43.800 --> 00:09:52.000
The next step--subtract: these cancel out; this is 0x² - 6x².
00:09:52.000 --> 00:09:59.300
And you can see how, if you didn't have that missing term in there, you would get a completely different answer.
00:09:59.300 --> 00:10:05.200
OK, so now I have -6x²; I am going to bring down this next term, 6x.
00:10:05.200 --> 00:10:19.700
OK, now x goes into -6x² -6x times; so I am going to put -6x up here.
00:10:19.700 --> 00:10:29.400
I divided; now multiply: -6x times x gives me -6x²; -6x times 2 is -12x.
00:10:29.400 --> 00:10:35.900
Now, I am subtracting; but remember, that is the same as adding the opposite--I am going to change these signs.
00:10:35.900 --> 00:10:44.400
These cancel out; and then, I end up with 6x + 12x, which is 18x.
00:10:44.400 --> 00:10:47.700
Bring down the next term, -4.
00:10:47.700 --> 00:10:58.800
x goes into 18x 18 times, so up here, I am going to have an 18.
00:10:58.800 --> 00:11:09.800
18 times x is 18x; 18 times 2 is 36; now I subtract.
00:11:09.800 --> 00:11:19.500
18x - 18x: that cancels; this negative applies to here as well, so I have -4 minus 36 to get -40.
00:11:19.500 --> 00:11:32.900
So, the remainder equals -40; and so, my answer is 3x² - 6x + 18, with a remainder of -40.
00:11:32.900 --> 00:11:38.700
Again, the key point here was to make sure that, when you notice that there is a missing term
00:11:38.700 --> 00:11:44.600
(the x² term is missing) to represent that using the coefficient of 0.
00:11:44.600 --> 00:11:49.300
OK, there is another way to divide that is much faster and easier than long division.
00:11:49.300 --> 00:11:55.000
But it is only applicable to dividing a polynomial by a binomial.
00:11:55.000 --> 00:12:01.200
So, when you have those cases, synthetic division is an excellent method to use.
00:12:01.200 --> 00:12:11.400
Now, there is another restriction, and that is that the divisor (which, in this case, is the binomial) must be in the form x - r, where r is a constant.
00:12:11.400 --> 00:12:18.500
So, the divisor has to be in this form; if it is not in that form, you have to get it into that form.
00:12:18.500 --> 00:12:26.900
OK, let's take a look at an example where the divisor is already in this form, to keep it simple for now.
00:12:26.900 --> 00:12:37.300
3x³ - 2x² + 7x + 4; if I am asked to divide that by x - 2,
00:12:37.300 --> 00:12:43.500
and it is in this form of x minus a constant, I could use long division.
00:12:43.500 --> 00:12:46.300
But you will see that this is a much faster method.
00:12:46.300 --> 00:12:57.100
So, first, draw a symbol like this; now, you put the r term right here (in this case, it is 2).
00:12:57.100 --> 00:13:05.600
And it says - 2, but you actually use the opposite sign; if this were to say x + 3, then I would put a negative here.
00:13:05.600 --> 00:13:11.800
So, this is x - 2; I use the opposite sign, so I am going to put a 2 right here.
00:13:11.800 --> 00:13:24.700
Now, what I write in here are the coefficients of the dividend: 3, -2, 7, and 4.
00:13:24.700 --> 00:13:30.400
And just as with long division, if there is a missing term (let's say my x² term was missing),
00:13:30.400 --> 00:13:35.300
I would have represented that here with a coefficient of 0--the same idea as with long division.
00:13:35.300 --> 00:13:41.400
OK, so here I have the constant; here I just have the coefficients from up here.
00:13:41.400 --> 00:13:46.400
The first step is to bring down that first coefficient and just put it here.
00:13:46.400 --> 00:13:59.400
OK, the second step is to multiply 2 times 3; I am going to multiply 2 times 3, and my result will be 6.
00:13:59.400 --> 00:14:09.500
All I am going to do is add this to -2; -2 + 6--that is going to give me 4.
00:14:09.500 --> 00:14:25.100
Then, I repeat that process: I am going to multiply -2 by 4 to get 8, and I am going to add: 8 + 7 is 15.
00:14:25.100 --> 00:14:35.900
I am going to repeat that again: -2 times 15 is going to give me 30; and I am going to add again: 4 + 30 is 34.
00:14:35.900 --> 00:14:48.000
Now, what do these numbers represent? Well, they represent the quotient and the remainder.
00:14:48.000 --> 00:14:52.100
And to write them out, what you are going to do is look at the dividend.
00:14:52.100 --> 00:14:57.200
And the quotient is going to be a degree one less than the degree of the dividend.
00:14:57.200 --> 00:15:01.600
So, the degree of the dividend was 3; so the degree of the quotient is going to be 2.
00:15:01.600 --> 00:15:15.500
I am going to write this out as 3x² + 4x + 15, with a remainder of 34.
00:15:15.500 --> 00:15:18.800
And you can see how much quicker and easier this is than long division.
00:15:18.800 --> 00:15:24.500
Again, write the constant from the divisor here; write the coefficients here, making sure to use 0
00:15:24.500 --> 00:15:29.600
if you have a missing term, using 0 as the coefficient; bring down the first term.
00:15:29.600 --> 00:15:37.300
Then, multiply the divisor constant by this number and get a product.
00:15:37.300 --> 00:15:46.600
Add that to the next column; then multiply 2 times this number, 4, to get 8; add to this column, 15.
00:15:46.600 --> 00:15:54.100
2 times 15 is 30; add to this column to get 34, which is the remainder.
00:15:54.100 --> 00:16:02.000
OK, as I mentioned, the divisor must be in the form x - r; if the coefficient of x is not 1,
00:16:02.000 --> 00:16:08.300
you need to rewrite it so that the coefficient is 1 in order to use this method.
00:16:08.300 --> 00:16:21.100
For example, if I was given 3x³ + 12x² - 15x + 6, divided by 3x - 9,
00:16:21.100 --> 00:16:27.000
I can see that this coefficient is not 1; in order to make it 1, I need to divide each term
00:16:27.000 --> 00:16:31.900
in both the dividend and the divisor by this coefficient.
00:16:31.900 --> 00:16:47.100
So, I am going to divide each term by 3; and that is going to give me x³ + 4x² - 15x + 2, divided by x - 3.
00:16:47.100 --> 00:16:50.200
Now, I can go about synthetic division in the usual way.
00:16:50.200 --> 00:16:56.600
And as you will see here, I got lucky, and dividing things by 3 still kept everything as integers.
00:16:56.600 --> 00:17:01.800
However, it is very possible that, when you divide by this number, you are going to end up with some fractions.
00:17:01.800 --> 00:17:10.300
So, that is a drawback, and it makes it more complicated; but it is necessary to do this in order to use synthetic division.
00:17:10.300 --> 00:17:14.700
OK, first we are going to practice dividing a polynomial by a monomial.
00:17:14.700 --> 00:17:22.600
And you will recall that the technique is to divide each term in the polynomial by the monomial.
00:17:22.600 --> 00:17:39.600
So, I am rewriting this, dividing each term by the monomial.
00:17:39.600 --> 00:17:53.100
So, 20 divided by 4 gives you 5; the x's cancel out; and then, y⁴ divided by y²...this is 4 - 2...is y², 5y².
00:17:53.100 --> 00:18:00.000
12 divided by 4 (and there is a negative sign in front of that) gives me 3.
00:18:00.000 --> 00:18:10.900
x to the first minus 2...this is going to give me x to the -1, and I am going to have to take care of that in a minute to put it in my final form.
00:18:10.900 --> 00:18:12.900
But for right now, we will leave it like that.
00:18:12.900 --> 00:18:23.300
y³ divided by y²...3 - 2 just gives me y, plus...I'll just leave this as 5/4...
00:18:23.300 --> 00:18:32.800
x³ divided by x² is x; and then here, I have y, which is really y¹,
00:18:32.800 --> 00:18:36.500
divided by y², is 1 - 2; and this is y^-1.
00:18:36.500 --> 00:18:42.500
Now, remember: you want to simplify things, and they are not in simplest form if you have negative powers.
00:18:42.500 --> 00:18:56.100
So, recalling that rule that a^-n equals 1/a^n, I can simplify by moving this x^-1 to the denominator.
00:18:56.100 --> 00:19:03.000
And I can do the same thing here: x stays up here; y moves to the denominator.
00:19:03.000 --> 00:19:09.800
And this is my answer: dividing a polynomial by a monomial, and then simplifying.
00:19:09.800 --> 00:19:19.400
OK, this next problem, dividing a polynomial by a monomial, could be done by synthetic division, because it is in this form x - r.
00:19:19.400 --> 00:19:31.900
But just to get a little more practice on long division, let's do long division on this one.
00:19:31.900 --> 00:19:47.100
x goes into 3x³...this becomes 2, so this is 3x²; so it goes into it 3x² times.
00:19:47.100 --> 00:19:52.800
So, I divided; and now I am going to multiply: 3x² times x is 3x³;
00:19:52.800 --> 00:19:57.100
3x² times -3 is -9x².
00:19:57.100 --> 00:20:05.200
Now, I am subtracting; and that is the same as adding the opposite, so this becomes negative; this becomes positive.
00:20:05.200 --> 00:20:12.700
These cancel out, so I have -2x² + 9x² is 7x².
00:20:12.700 --> 00:20:16.800
Bring down the next term, and then divide again.
00:20:16.800 --> 00:20:32.700
I have 7x² divided by x equals 7x; 7x times x; 7x, -3 gives me -21x.
00:20:32.700 --> 00:20:40.500
Subtract (which means I am adding the opposite): this becomes a negative; this one becomes a positive.
00:20:40.500 --> 00:20:55.000
This cancels; 4x + 21x is 25x; now, divide again: 25x divided by x...the x's cancel; that is 25.
00:20:55.000 --> 00:20:57.400
So, I am going to have 25 up here.
00:20:57.400 --> 00:21:09.500
Multiply 25 times x; that is 25x; and bring down this next term; I have -8 there.
00:21:09.500 --> 00:21:20.600
So, 25 times -3 is going to give me -75; I am going to subtract, which is adding the opposite.
00:21:20.600 --> 00:21:28.800
So, that is -25 and positive 75; these cancel; -8 + 75 is 67, and that is my remainder.
00:21:28.800 --> 00:21:38.100
So, the answer is 3x² + 7x + 25, with a remainder of 67.
00:21:38.100 --> 00:21:47.400
That was long division; now, this next example specifies to divide using synthetic division.
00:21:47.400 --> 00:22:01.000
And I am checking, and the divisor is in the form x - r; so I can go ahead and do it without any further manipulation of the expression.
00:22:01.000 --> 00:22:08.700
Recall: in synthetic division, set it up as follows: you are going to put the constant here, with the opposite sign.
00:22:08.700 --> 00:22:11.800
This is -4; I am going to make it a 4.
00:22:11.800 --> 00:22:18.800
Now, before I proceed with putting the coefficients in, it is very important to check for missing coefficients.
00:22:18.800 --> 00:22:27.500
And I am looking, and I have y⁴; I do not have y³; I have y², y, and a constant.
00:22:27.500 --> 00:22:37.900
So, I am going to rewrite this using 0 for my coefficient with the missing term.
00:22:37.900 --> 00:22:48.900
I have a missing y³ term, and I am going to rewrite it; and the coefficient is 0 (to use that as a placeholder).
00:22:48.900 --> 00:22:52.200
So, this is really what I am going to do; OK.
00:22:52.200 --> 00:23:06.100
Now, I am going to use these coefficients: I have 5, 0, -3, 2, and then my constant, -8.
00:23:06.100 --> 00:23:12.300
The first step is just to bring down that first term, 5; the second step is to multiply.
00:23:12.300 --> 00:23:37.000
4 times 5 is 20; after multiplying, add: 0 and 20 is 20; multiply again: 4 times 20 is 80; -3 and 80 is 77.
00:23:37.000 --> 00:23:51.300
4 times 77--if you calculated that out, you would find that it is 308; 308 + 2 is 310.
00:23:51.300 --> 00:24:03.700
4 times 310--if you work that out, you will find that it is 1240; 1240 and -8 is going to give you 1232.
00:24:03.700 --> 00:24:08.500
So again, write the coefficients here, being careful to realize you have a missing term,
00:24:08.500 --> 00:24:12.000
so you need to represent the coefficient for that missing term as 0.
00:24:12.000 --> 00:24:19.900
Bring down the first term, and then multiply and add that product to the next column.
00:24:19.900 --> 00:24:26.800
Find the sum; multiply; add the product to the next column; find the sum; and continue on.
00:24:26.800 --> 00:24:39.600
Now, for my quotient: this is the quotient; this is going to be the remainder.
00:24:39.600 --> 00:24:43.200
The quotient is going to have a degree one less than the degree of the dividend;
00:24:43.200 --> 00:24:47.800
so the degree of the dividend is 4, so the degree of the quotient is going to be 3;
00:24:47.800 --> 00:24:57.900
so I am going to write this out as 5y³ + 20y² + 77y + 310;
00:24:57.900 --> 00:25:09.900
the remainder is 1232, which is a big remainder, but this is correct.
00:25:09.900 --> 00:25:17.900
In this example, we also are asked to do synthetic division; and I check, and I have a couple of things going on here.
00:25:17.900 --> 00:25:26.000
I have a missing coefficient, and this is not in the correct form of x - r.
00:25:26.000 --> 00:25:33.400
So, dealing with the missing term--I have a missing term, which I am going to represent with a coefficient of 0.
00:25:33.400 --> 00:25:42.800
So, first addressing the missing term: I am missing a z² term; I have 6z⁴ - 8z³.
00:25:42.800 --> 00:25:52.400
Since I have no z² term, I am going to use the coefficient of 0 for that: 0z² - 4z + 8.
00:25:52.400 --> 00:25:58.500
OK, that is taken care of; now, the other problem I have is that this is not in the correct form.
00:25:58.500 --> 00:26:09.500
In order to have this z have a coefficient of 1, I need to divide all the terms in the divisor and the dividend by 2, so I am going to do that.
00:26:09.500 --> 00:26:29.700
So, I divide by 2--divide each term by 2 to get the form...I am going to say z instead of x...the form z - r.
00:26:29.700 --> 00:26:36.000
OK, so 6z⁴: dividing that by 2 is going to give me 3z⁴.
00:26:36.000 --> 00:26:51.600
8z³ divided by 2 is -4z³; 0z² divided by 2 is 0z².
00:26:51.600 --> 00:27:00.600
-4z divided by 2 is -2z; 8 divided by 2 is 4; so far, pretty good.
00:27:00.600 --> 00:27:08.100
2z divided by 2 gives me z, which is what I wanted; now, here, when I divide -1 by 2, I am going to get a fraction.
00:27:08.100 --> 00:27:14.400
And it makes it more difficult to work with, but you can still do the synthetic division.
00:27:14.400 --> 00:27:24.400
Now, I am ready to set this up: here, I am going to take -1/2; I am going to take the opposite sign and write it here; that is 1/2.
00:27:24.400 --> 00:27:39.400
Then, I am going to put the coefficients here--my new coefficients, after dividing: 3, -4, 0, -2, and 4.
00:27:39.400 --> 00:27:50.200
OK, bring down the 3 and multiply 3 by 1/2; 3 times 1/2 is just 3/2.
00:27:50.200 --> 00:27:56.800
1/2 times 3 is 3/2; now, you might need to work out the arithmetic on the side, and that is fine.
00:27:56.800 --> 00:28:04.900
I have -4 and 3/2; -4 is equal to -8/2; I want to get a common denominator.
00:28:04.900 --> 00:28:15.600
Adding that to 3/2 is going to give me -5/2, so this is -5/2.
00:28:15.600 --> 00:28:29.200
Now, multiplying 1/2 by -5/2 is going to give me -5/4; this times this is going to give me -5/4.
00:28:29.200 --> 00:28:44.600
0 and -5/4 is just -5/4; now, I have to multiply 1/2 by -5/4, and this is going to give me -5/8; I have -5/8 here.
00:28:44.600 --> 00:28:56.700
I want to get a common denominator; and so, I am going to do -2 times 8, which is going to give me -16/8.
00:28:56.700 --> 00:29:13.100
So, this is -16/8 + -5/8; and it is just -16 and -5, is -21/8; I am combining these to get -21/8.
00:29:13.100 --> 00:29:31.100
OK, now 1/2 times -21/8 equals -21/16; this times this is -21/16.
00:29:31.100 --> 00:29:48.200
I have to add that to 4; so I want to get a common denominator: 4 times 16 is going to give me 64, so 4 = 64/16.
00:29:48.200 --> 00:30:01.100
So, I want to take 64/16 - 21/16; and that is going to be...64 - 21 is 43/16.
00:30:01.100 --> 00:30:09.400
OK, this was kind of messy to do; but synthetic division does work, and will end up giving you the correct answer.
00:30:09.400 --> 00:30:16.000
So, what you need to look at here is that this has a degree of 4, so this is actually going to be a degree of 3.
00:30:16.000 --> 00:30:41.600
So, rewriting this up here, my quotient is going to be 3z³ - 5/2z² - 5/4z - 21/8.
00:30:41.600 --> 00:30:45.900
And I have a remainder of 43/16.
00:30:45.900 --> 00:30:51.900
So, two things to notice: we had a missing term here--the z² term was missing--
00:30:51.900 --> 00:30:56.000
so I had to use a coefficient of 0; and the second is that this was not in this form.
00:30:56.000 --> 00:31:05.200
So, I had to divide every term in the numerator and the denominator, the divisor and the dividend, by 2, and then proceed as usual with synthetic division.
00:31:05.200 --> 00:31:10.700
That concludes this lesson on dividing polynomials for Educator.com.
00:31:10.700 --> 00:31:11.000
And I will see you again soon!