Dividing Polynomials
 When dividing a polynomial by a monomial, think of dividing each term by the monomial.
 If we are dividing a polynomial by another polynomial, we can use long division. Before using this process, make sure both polynomials are written with descending powers, and with zero placeholders for missing powers.
 The long division process for polynomials is similar to the long division process of numbers. Donâ€™t forget to subtract away the entire polynomial at each step. You will often have to distribute a negative sign to do so.
 When dividing a polynomial by a polynomial of the form x â€“ k, you can use synthetic division to make the process cleaner. In this process we only write down the coefficients of each of the polynomials.
 Remember to add and multiply carefully when using synthetic division, and that the last number given is the remainder.
Dividing Polynomials
[(12x^{2} − 4x + 10x − 6)/2x]
 [(12x^{3})/2x] − [(4x^{2})/2x] + [10x/2x] − [6/2x]
[(25y^{4} + 60y^{3} − 75y)/5y]
[(54m^{4} + 60m^{2} − 48m + 24)/6m]
[(2n^{2} + 3n − 20)/(n + 4)]
 [(( n + 4 )( 2n − 5 ))/(n + 4)]
[(3n^{2} − n − 24)/(n − 3)]
 [(( 3n + 8 )( n − 3 ))/(n − 3)]
[(( 5j^{2} + 14 )( j + 2 ))/(j + 2)]
 [(( 3p + 10 )( 2p − 1 ))/(3p + 10)]
[(( 4x^{3} − 10x^{2} + 12x − 8 ))/(( x − 2 ))]
 [((4x^{3}−10x^{2} +12x−8))/((x−2))]

4x^{2} x−2 ) 4x^{3} −10x^{2} +12x −8 −(4x^{3} −8x^{2}) −2x^{2} 
4x^{2} x−2 ) 4x^{3} −10x^{2} +12x −8 −(4x^{3} −8x^{2}) −2x^{2} +12x −(2x^{2} +4x) 8x 
4x^{2} x−2 ) 4x^{3} −10x^{2} +12x −8 −(4x^{3} −8x^{2}) −2x^{2} +12x −(2x^{2} +4x) 8x −8 −(8x −16) 8  (x−2)(4x^{2}−2x+8)+8
 4x^{3} −2x^{2}+8x−8x^{2}+4x−16+8
[(( 5y^{3} + 35y − 10y + 65 ))/(( y + 5 ))]

y+5 ) 5y^{3} 35y^{2} −10y +65 
5y^{2} y+5 ) 5y^{3} 35y^{2} −10y +65 −(5y^{3} +25y^{2}) 10y^{2} 
5y^{2} +10y y+5 ) 5y^{3} 35y^{2} −10y +65 −(5y^{3} +25y^{2}) 10y^{2} −10y −(10y^{2} +50y) −60y 
5y^{2} +10y −60 y+5 ) 5y^{3} 35y^{2} −10y +65 −(5y^{3} +25y^{2}) 10y^{2} −10y −(10y^{2} +50y) −60y +65 −(−60y −300) 365  (y+5)(5y^{2}+10y−60)+365
 5y^{3}+10y^{2}−60y+25y^{2}+50y−300+365
[((6a^{2}+12a^{2}−11a+3))/(a+1)]

6a^{2} 3a+1 ) 6a^{2} 12a^{2} −11a +3 −(6a^{3} +6a^{2}) 6a^{2} 
6a^{2} −2a 3a+1 ) 6a^{2} 12a^{2} −11a +3 −(6a^{3} +6a^{2}) 6a^{2} −(−6a^{2} −2a) −9a 
6a^{2} −2a −3 3a+1 ) 6a^{2} 12a^{2} −11a +3 −(6a^{3} +6a^{2}) 6a^{2} −(−6a^{2} −2a) −9a −(−9a −3) 6  (3a+1)(6a^{2}−2a−3)+6
 18a^{3}−6a^{2}−9a+6a^{2}−2a−3+6
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Dividing Polynomials
Lecture Slides are screencaptured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
 Intro 0:00
 Objectives 0:07
 Dividing Polynomials 0:29
 Dividing Polynomials by Monomials
 Dividing Polynomials by Polynomials
 Dividing Numbers
 Dividing Polynomials Example
 Example 1 12:35
 Example 2 14:40
 Example 3 16:45
 Example 4 21:13
 Example 5 24:33
 Example 6 29:02
 Dividing Polynomials with Synthetic Division Method
 Example 7 38:43
 Example 8 42:24
Algebra 1 Online Course
Transcription: Dividing Polynomials
Welcome back to www.educator.com.0000
In this lesson we are going to take care of dividing polynomials. 0003
I like to break this down into two different parts. 0009
First, we will look at the division process when you have a polynomial divided by a monomial.0013
We will look at what happens when you divide any two polynomials together.0017
In the very end, I will also show you a very special technique to make the division process nice and clean. 0022
To get into the basics of understanding the division of polynomials, I like to take it back to looking at the division process for realworld fractions. 0031
Suppose that when you were adding fractions together you know how that process will go.0042
One very important thing that you do when adding fractions is you would find a common denominator. 0047
Once you have a common denominator, then you can go ahead and just combine the tops of those fractions together.0053
Think of a quick example like 2/3 and are looking to add 5/3.0060
Since they have exactly the same bottoms then you will only add the 2 and 5 together and get 7/3 as your result.0067
Since this has a giant equal sign in between it, it means you can also follow this process in the other direction.0074
This may look a little unfamiliar, but it does work out if you go the other way. 0080
Suppose I had A + B and all of that was dividing by C.0085
The way I could look at this is that both the A and B are being divided by C separately.0090
This is the same equation I had earlier, I just turned it around. 0097
The reason why I show this is this will help us understand what happens when we have a binomial divided by a monomial.0102
In fact, that is the example that I have written out.0108
The top is a binomial and the bottom is an example of a monomial.0112
You can see that the way we handle it is we split up that monomial under each of the different parts of the polynomial in the top.0121
Let us see this process with numbers and see what polynomials and you will get an idea of how this works.0134
Working the other direction, if I see (2 + 5) Ã· 3, I want to visualize that as the 2 Ã· 3 and also the 5 Ã· 3.0140
For our polynomials if I have something like (x + 3z) Ã· 2y, then I will put that 2y under both of the parts.0152
In addition, you will notice splitting up over both of the parts in the top, you always want to make sure that you simplify further, if possible. 0162
This means if you use your quotient rule for exponents, go ahead and do reduce those powers as much as possible. 0172
If you have just any two general polynomials, then you want to think of how the process works with numbers.0183
In fact we are going to go over a long division process so that we can actually keep track of all the parts of what goes into what.0189
To make this process easier, remember to write your polynomial in descending power.0199
Start with the largest power and write it all the way down to the smallest power. 0203
An additional thing that will also help in the division process is to make sure you put placeholders for all the missing variables.0214
If Iâ€™m looking at a polynomial like 5x^{2} + 1, then I will end up writing it with a placeholder for the missing x.0221
It is not missing but it will help me keep track of where just my x terms go.0233
I'm going to show you this process a little bit later on0239
so you can see how it works with numbers and make some good parallels too doing this with polynomials.0242
Let us look at the division process for numbers.0251
Suppose I gave you 8494 and I told you to divide it by 3 and furthermore I said okay, let us see if you can do this by hand.0254
Some good news that you will probably tell me is that you do not have to take the 3 into that entire number all at once. 0263
No, you will just take the 3 into 8494 bit by bit.0269
In fact, the first thing that you will look at is how many times this 3 go into the number 8.0274
That will be the only thing you are worried about.0279
How many times does the 3 go into 8?0282
It goes in there twice and you will write that number on the top. 0285
Now after you have that number on the top, you do not leave it up there you go through a multiplication process, 0290
2 Ã— 3 and you will write the result right underneath the 8.0296
With the new number on the bottom, you will go ahead and subtract it away, so 86 would give you a 2. 0305
That would be like one step of the whole division process.0315
You would continue on with the division process by bringing down more terms and doing the process again.0319
At this next stage, we will say okay how many times this 3 go into 24?0329
It goes in there 8 times.0335
Then we could multiply the 8 and 3 together and get a new number for that 3 Ã— 8 = 24.0338
We can go ahead and subtract those away.0352
We will not stop there, keep bringing down our other terms and see how many times 3 goes into 9.0359
Write it onto the top and multiply it by your number out front.0369
Subtract it away and continue the process until you have exhausted the number you are trying to divide.0377
Let us see.0389
3 goes into 4, it looks like it goes in there once.0390
I will get 3, subtract them away and I have remainder of 1.0396
That is a lengthy process but notice the Q components in there.0401
You are only dividing the number bit by bit.0405
You do not have to take care of it all at once.0408
The way you take care of it is you are saying how many times 3 goes into that leading number.0410
You write it on the top, you go through the multiplication process and then you subtract it away from the number.0415
You will see all of those same components when we get into polynomials.0423
We have our answer and I could say it in many different ways but Iâ€™m going to write it out.0429
8494 if we divide this by 3 is equal to 2831 and it has a remainder down here of 1.0434
We could say +1 and still being divided by 3.0446
We have many different parts in here that are flying around, and you want to keep track of what these parts are.0451
The part underneath your division bar is the dividend. 0458
What you are throwing in there this is your divisor.0466
Your answer would be your quotient.0472
This guy down here is our remainder. 0477
Notice how those same parts actually show up in our answer.0483
Dividend, divisor, quotient, and remainder.0488
We put the remainder over the divisor because it is still being divided.0506
Now that we brushed up on the process with numbers, let us take a look at how we do this with polynomials. 0510
I want to divide 2x^{2} + 10x + 12 and divide that by x +3.0519
The good news is we do not have to take care of the entire polynomial all at once.0526
We are going to take it in bits and pieces.0530
We will first going to look at x and see how many times it will go into 2x^{2}.0532
In fact one thing that I can do to help out the process is think to myself what would I have to multiply x by in order to get a 2x^{2}.0537
1 Ã— what would equal to 2? That would have to be 2.0548
What would I have to multiply x by to get an x^{2}?0552
I have to multiply it by x.0555
I will put that on top, just like I did with numbers.0557
After I do that, we will run through a multiplication process.0563
We will take the 2x to multiply it by x, I will multiply it by 3.0568
We will record this new polynomial right underneath the other one.0573
2x Ã— x = 2x^{2}.0577
2x Ã— 3 = 6x.0581
Now comes a very important step.0589
Now that we have this new one, we want to subtract it away from the original.0592
Notice how I put those parentheses on there, that will help me remember that I need to subtract away both parts and keep my signs straight.0599
Starting over here, I have 10x â€“ 6x = 4x.0606
Then I have 2x^{2} â€“ 2x^{2}, that will cancel out and will give me 0x^{2}.0616
If you do this process correctly, these should always cancel out.0622
If they do not cancel out, it means we need to choose a new number up here.0628
That is just one step of the division process, let us bring down our other terms and try this one more time.0635
I want to figure out how many does x goes into 4x?0648
What would I have to multiply x by in order to get 4x?0652
I think I have to multiply it by 4 that is the only way it is going to work out.0658
Now we have the 4, go ahead and multiply it by the numbers out front.0663
4 Ã— x = 4x and 4 Ã— 3 = 12.0669
Once you have them, put on a giant pair of parenthesis and we will go ahead and subtract it away.0678
12 â€“ 12 =0 and 4x â€“ 4x = 0.0687
What that shows is that there is no remainder and that it went evenly.0693
Let us write this out.0699
When I had 2x^{2} + 10x + 12 and I divided it by x + 3, the result was 2x + 4.0699
There was no remainder.0714
We can label these parts as well.0717
We have our dividend, divisor, quotient, and if I did have a remainder I will probably put it out here.0720
Here is our dividend.0739
Here is our divisor and our quotient.0744
Now that we know a lot more about dividing polynomials, let us look at a bunch of examples.0758
Some of them will take a polynomial divided by a monomial.0764
Some of them will take two polynomials and divide them.0767
We will approach both of those cases in two different ways.0770
Example 1, divide a polynomial by a monomial.0774
I will take (50m^{4} 30m^{3} + 20m) Ã· 10m^{3}.0777
Since we are dividing by a monomial, I will take each of my terms and put them over what Iâ€™m dividing them by.0784
50m^{4} Ã· 10m^{3}, 30m^{3} Ã·10m^{3} and 20m Ã· 10m^{3}.0792
Now that I have done that, I will go through and simplify these one at a time.0811
50 Ã· 10 = 5, m^{4} Ã· m^{3} = I can use my quotient rule and simply subtract the exponents and get m^{1}.0816
Continuing on 30 Ã· 10 = 3, if I subtract my exponents for m^{3} and m^{3}, I have m^{0}.0830
Onto the last one, 20m Ã· 10m^{3}, there is 2.0844
Let us see, if I subtract the exponents I will m^{2}.0852
Then I can go through and just clean this up a little bit.0857
5m â€“ anything to the 0 power is 1, 1 Ã— 3 + and I will write this using positive exponents, 2/m^{2}.0860
This will be the final result of dividing my polynomial by a monomial.0872
Let us try this scheme with something a little bit more complicated. 0881
This one is (45x^{4} y^{3} + 30x^{3} y^{2}  60x^{2} y) Ã· 50x^{2} y.0884
We have to put our thinking caps for this one.0894
We will start off by taking all of our terms in the top polynomial and putting them over our monomial, only one term.0897
Once we have this all written out we simply have to simplify them one at a time.0917
Let us start at the very beginning.0925
15 goes into 45 3 times, now I will simplify each of my variables using the quotient rule.0929
4 â€“ 2 =2 and 3 â€“ 1= y^{2}.0939
There is my first term, moving on.0946
15 goes into 30 twice.0950
Using my quotient rule on the x, 3 â€“ 2 = 1 and y^{2} â€“ y^{1} = y^{1}.0955
Both of these have an exponent of 1 and I do not need to write it.0964
One more, 15 goes into 60 four times.0969
I have x^{2}/x^{2} which will be x^{0} and y/y will be y^{0}.0975
Anything to the 0 power is 1.0983
I can end up rewriting that term 3x2 y^{2} + 2xy â€“ 4.0987
There is my resulting polynomial.1000
In this next example, we are going to take one polynomial divided by another polynomial.1007
I would not be able to split it up quite like I did before, now we will have to go through that long division process.1012
Be careful you want to make sure that you line up your terms in descending order. 1018
If you look at the powers of the top polynomial you would not mix up.1024
I want to start with that 3rd power then go to the 2nd power, then go to the 1st power, just to make sure I have it all lined up.1027
Let us write it out.1034
2x^{3} + x^{2} + 5x + 13, now it is in a much better order to take care of.1035
We will take all of that and we will divide it by 2x + 3.1048
That looks good.1055
It is time to get into the division process.1057
Our first terms there, and what would I need to multiply 2x by in order to get 2x^{3}?1060
The only thing that will work would be an x^{2}.1068
Once we find our numbers up top, we will multiply them by the polynomial out front.1072
2x Ã— x^{2} = 2x^{3}, that is a good sign, it is the same as the number above it.1079
X^{2} Ã— 3 =3x^{2}, now comes the part that is tricky to remember.1088
Always subtract this away and do not be afraid to use this parenthesis to help you remember that.1097
x^{2} â€“ 3x^{2}, 1 â€“ 3 = 2x^{2}.1106
Then I have 2x^{3} â€“ 2x^{3} and those will be gone, cancel out like they should.1116
The first iteration of this thing looks pretty good. 1122
Let us try another one.1126
I want to figure out how may times does 2x go into 2x^{2}?1129
I have to multiply it by â€“x.1137
Let us bring down some more terms.1146
I will get onto our multiplication process.1149
x Ã— 2x =2x^{2}.1153
x Ã— 3 =3x.1158
We have our terms in there, it is time to subtract it away and be careful with all of our signs.1164
5x  3x when we subtract a negative that is the same as addition.1171
Iâ€™m looking at 5x + 3x = 8x.1179
2x^{2}  2x^{2} = that is a lot of minus signs.1188
That would be the same as 2x + 2x^{2} and they do cancel out like they should.1193
That was pretty tricky keeping track of all those signs but we did just fine.1201
We have 8x + 13 and Iâ€™m trying to figure out what would I have to multiply 2x by in order to get that 8x.1207
There is only one thing I can do I need to multiply by 4.1218
We will multiply everything through and see what we get.1224
4 Ã— 2x = 8x, 4 Ã— 3 =12.1228
I can subtract this away and let us see what we get.1238
13 â€“ 12=1, 8x â€“ 8x = 0.1245
Here I have a remainder of 1.1250
I could end up writing my quotient and I can put my remainder over the divisor.1257
There is our answer.1269
In some polynomials you want to make sure you put in those placeholders.1275
That way everything lines up and works out good.1280
It is especially what we will have to do for some of those missing powers in this one. 1283
Iâ€™m missing an x^{2} and a single x.1287
Let us write this side and put those in.1291
x^{3}, I have no x^{2}, no x â€“ 8.1294
All of that is being divided by x â€“ 2.1305
Iâ€™m going to go through and let us see what I need to multiply x by in order to get x^{3}.1310
I think Iâ€™m going to need x^{2}.1318
Now that I found it, I will go ahead and multiply it through.1323
x^{2} Ã— x =x^{3} and x^{2} Ã— 2 = 2x^{2}.1327
We have our terms, let us go ahead and subtract it away.1336
0x^{2} â€“ 2x^{2}, this is one of those situations where if we subtract a negative is the same as adding.1345
0x^{2} + 2x^{2} =2x^{2}.1354
x^{3} â€“ x^{3}, that is completely gone.1360
Bring down our other term here and we will keep going.1366
What would I have to multiply x by in order to get 2x^{2}?1374
Iâ€™m going to need 2x.1381
Let us multiply that through, 2x Ã— x =2x^{2}.1386
2x Ã— 2 =4x.1392
Now we found that, let us subtract that away.1399
Starting on the end, 0 â€“ 4, that is the same as 0 + 4x = 4x.1405
Then 2x^{2} â€“ 2x^{2}, they are completely gone, you do not have to worry about it.1415
We will bring down our 8 and continue.1422
What would I have to multiply x by in order to get 4x?1428
That will have to be 4.1433
4 Ã— x = 4x and 4 Ã— 2 = 8.1437
It looks like it is exactly the same as the polynomial above it.1447
I know when I subtract, I would get 0.1452
There is no remainder for this one.1454
I will take (x^{3} â€“ 8) Ã· x 2 and the result is x^{2} + 2x + 4.1457
Division process can take a bit but as long as you do the steps very carefully, you should turn out okay.1467
Let us try this giant one.1476
This is (2m^{5} + m^{4} + 6m^{3} â€“ 3m^{2} â€“ 18) Ã· m^{2} + 3.1478
There are a lot of things to consider in here.1489
One thing that I will be careful of is putting those placeholders for this guy down here.1493
Notice that it is missing an m, let us give it a try.1498
I will have m^{2} + 0m + 3 and all of that is going into our other polynomial 2m^{5} + m^{4} + 6m^{3} â€“ 3m^{2} â€“ 18.1501
Lots of things to keep track of but I think we will be okay.1523
What would I have to multiply the m^{2} by in order to get a 2m^{5}?1527
That would be 2m^{3}.1535
I can run through the multiplication and write it here.1539
2m^{3} Ã— m^{2} = 2m^{5}.1543
Now we can multiply it by our 0 placeholder but anything times 0 will give us 0 so 0m^{4}.1549
By my last one, let us see 2m^{3}Ã— 3 + 6m^{3}.1558
Let us subtract that away.1568
6m^{3} â€“ 6m^{3} those are gone.1572
I have m^{4} â€“ 0m^{4} I still have m^{4}.1575
2m^{5} â€“ 2m^{5} those are gone.1580
I dropped away quite a bit of terms.1584
Let me go ahead and write in my 0m^{3} as one of those placeholders so I can keep track of it.1587
Let us try this again.1598
m^{2} goes into m^{4}, if I multiply it by another m^{2}.1600
Multiplying through I have m^{2} Ã— m^{2} = m^{4}.1609
m^{2} Ã— 0m = 0m^{3} and m^{2} Ã— 3 = 3m^{2}.1614
We will take that and subtract it away.1626
I need to go ahead and subtract these.1636
Be very careful on the signs of this one.1638
I have â€“m^{2} and Iâ€™m subtracting 3m^{2}, the result here will be 6m^{2}.1641
The reason why it is happening is because of that negative sign out there.1652
Now I have 0m^{3} â€“ 0m^{3}, 0 â€“ 0 =0, m^{4} â€“ m^{4} = 0.1659
It looks like I forgot an extra placeholder.1669
I need 0m and then I need my 18.1676
Let us bring down both of these.1684
I need to figure out what I have to what would I have to multiply m^{2} by in order to get 6m^{2}.1691
6 will do it, 6m^{2} 6 Ã— 0 = 0m and 6 Ã— 3 =18.1699
It is exactly the same as the polynomial above it.1718
Since they are exactly the same and Iâ€™m subtracting one from the other one, 0 is the answer.1723
There is no remainder, it went evenly.1729
The quotient for this one would be 2m^{3} + m^{2} â€“ 6.1732
In this polynomial, I have (3x^{3} + 7x^{2} + 7x + 11) Ã· 3x + 6.1744
The reason why I put this one is because it can get a little bit difficult figuring out what you need to multiply to get into that second polynomial.1753
Iâ€™m going to warn you, this involves a few fractions.1762
Let us give it a shot.1767
(3x^{3} + 7x^{2} + 7x + 11) Ã· 3x + 6.1770
Let us start off at the very beginning.1788
What do I need to multiply my 3x by in order to get 3x^{3}?1790
The only thing that will work will be an x^{2}.1797
I will go through and I will multiply and get the result.1804
3x^{3} + 6x^{2} and let us subtract that away.1809
7x^{2} â€“ 6x^{2} = 1x^{2}.1821
Now comes the tricky part, I need to figure out what I need to multiply 3x by in order to get x^{2}.1837
If Iâ€™m looking at just the variable part of this, I have to multiply and x by another x in order to get an x^{2}.1845
We will go ahead and put that as part of our quotient.1851
What do I have to multiply 3 by in order to get 1 out front?1854
That is a little bit trickier.1859
3 Ã— what = 1.1861
That is almost like an equation onto itself.1865
What we see is that x would have 1/3, a fraction.1869
It is okay, we can use fractions and end up multiplying by those.1874
3x Ã— 1/3x = 1x^{2}.1881
Let us multiply that through.1887
1/3x Ã— 3x = 1x^{2} I will write it down and 1/3x Ã— 6 = 2x.1888
Now we can take that and subtract it away.1900
7x â€“ 2x = 5x.1904
Bringing down our extra terms and I think this one is almost done.1913
What would I have to multiply 3x to get 5x?1917
Let us see.1922
he only way I can get an x into another x is to multiply by 1, but Iâ€™m going to think of how do I get 3 and turn it into 5?1924
Let us do a little bit of scratch work on this one.1933
3 Ã— what = 5?1939
If we divide both sides by 3 I think we can figure out it is 5/3.1945
That I can write on top 5/3.1951
We can go through multiplying.1956
5/3 Ã— 3 = 5x.1958
5/3 Ã— 6 =10.1966
We will go ahead and subtract this away.1977
11 â€“ 10 = 1 and 5x â€“ 5x = 0.1980
We have a remainder of 1.1985
Now that we have all of the quotient and the remainder, let us go ahead and write it out.1993
We have (x^{2} + 1/3x + 5/3 + 1) Ã· 3x + 6.1998
Definitely do not be afraid some of those fractions to make sure it goes into that second polynomial.2010
This process can get a little messy as you can definitely see from those examples.2018
I have a nice clean way that you can go through the division process known as synthetic division.2022
This is a much cleaner way for the division process so that you can keep track of all the variables.2027
It is much clean but be very careful in how you approach this. 2033
It works good when dividing by polynomials of the form x + or â€“ number.2037
It will work especially with my little example right here (5x^{3}  6x^{2} +8) Ã· x 4.2043
First watch how I will set this up. 2051
I'm going to create like a little upside down division bar and that is where I will end up putting the polynomial that I'm dividing.2054
But I will not put the entire thing Iâ€™m only going to put the coefficients of all of the terms.2062
The coefficient of the x^{3} is a 5. 2069
The coefficient of my x^{2} is 6.2073
I will put a 0 placeholder in for my missing x and then my last coefficient will be 8.2077
Once I have all of those I will put another little line. 2086
I want to put in the value of x that would make this entire polynomial 0, if x was 4 that would be 0.2090
Iâ€™m going to write 4 out here.2105
That turns to be a tricky issue and many students remember what to put out over here because it will be the opposite of this one.2109
If you see x â€“ 4 put in a 4.2118
If you see something like x + 7 then put in 7.2121
We got that all set let us go through this synthetic division process.2127
It tends to be quick watch very carefully how this works.2131
The very first thing that you do in the synthetic division process is you take the first number here and you simply copy it down below.2137
This will be a 5.2146
Once you get that new number on the bottom, go ahead and multiply it by your number out front, 4Ã—5 = 20.2149
That is one step of the synthetic division process.2161
To continue from there simply add the column 6 + 20 and get the result. 2165
This would be a 14.2173
Once you have that feel free to multiply it out front again.2177
14 Ã— 4 = 56.2181
There are 2 steps now we will take the 56 and we will add 0, 56.2191
When we get our new number on the bottom, go ahead and multiply it right out front.2200
4 Ã— 50 = 200.2205
4 Ã— 6 = 24, 224.2212
One last part to this we got to do some addition.2222
8 + 224 = 232.2225
It does not look like we did much of any type of division. 2234
We did a lot of adding and we did a lot of multiplying but do you know what these new numbers stand for on the bottom.2236
That is the neat part, these new numbers I have here in green stand for the coefficients of our result.2243
You know what happens after the division.2249
The way you interpret these is the last number in this list will always be your remainder.2252
I know that my remainder is 232.2260
As for the rest of the values, the 5, 14, and 56, those are the coefficients on our variables. 2264
What should they be? Let me show you how you can figure that out.2270
Originally we had x^{3} as the polynomial that we are dividing and these new ones will be exactly one less in power.2274
That 5 goes with 5x^{2} and the 14 goes with the 14x and 56 has no x on it. 2282
The result for this one is 5x^{2} + 14x + 56 with a remainder of 232 which you can write over x â€“ 4.2293
Since it is still being divided2312
It is a much cleaner and faster method for division.2315
Let us go ahead and practice it a few times just to make sure got it down.2318
We will go ahead and do (10x^{4}  50x^{3} â€“ 800) Ã· x â€“ 6.2325
It is quite a large problem.2332
We will definitely remember to put in some of those placeholders to keep track of everything.2334
First I will write in all of the coefficients of my original polynomial.2339
I have a 10x^{4}  50x^{3} I need to put in a placeholder for my x^{2} and another placeholder for my x.2344
We will go ahead and put in that 800.2356
What shall we put on the other side?2364
Since I'm dividing by x â€“ 6, the value of 6 will be divided that would make that 0.2367
I will use 6 and now onto the synthetic division process.2376
The first part we will drop down to 10, just as it is.2381
Then we will multiply by the 6 out front 60.2387
Now that we have that, let us add the 50 and the 60 together, 10 again.2396
We will multiply this out front.2405
That result will be 60.2409
We will go ahead and add 0 + 60 = 60 and multiply 60 Ã— 6= 360.2417
Now we will add 0 + 360 = 360.2430
I have to take 360 Ã— 6.2435
think I have to do a little bit of scratch work for that one.2442
6 Ã— 0 = 6 Ã— 6 = 36 and then 3 Ã— 6 = 18 + 3 = 21.2443
I have 2160, let us put it in.2455
Only one last thing to do is we need to add 800 to the 2160 and then let us do a little bit of scratch work to take care of that.2465
0 â€“ 0= 0, 6 â€“ 0= 6, 20 and we will breakdown and will say 11 â€“ 8 =3.2478
I have 1, I have 1360.2489
Now comes the fun part, we have to interpret exactly what this means.2493
Keep in mind that this last one out here that is our remainder.2498
Originally our polynomial was x^{4} so we will start with x^{3} in our result.2505
10 x^{3} + 10x^{2} + 60x + 360 and then we have our remainder 1360.2513
It is all still being divided by an x â€“ 6.2532
That was quite a bit of work, but it was a lot clean than going through the long division process. 2536
Let us see this one more time.2540
In this last one we will use (5  3x + 2x^{2} â€“ x^{3}) Ã· x + 1.2545
We will first go ahead and put the coefficients of our top polynomial in descending order.2555
Be very careful as you set this one up.2560
On that side you can write out the polynomial first in descending order and then go ahead and grab its coefficients.2564
x^{3} would be the largest, then I have 2x^{2} and then I have 3x, and 5.2570
I need 1, 2, 3 and 5.2579
Iâ€™m dividing by x + 1 so the number I will use off to the left will be 1.2589
I think we have it all set up now let us run through that process.2597
The first Iâ€™m going to bring down is 1 then we will go ahead and multiply.2600
Negative Ã— negative is positive.2608
2 + 1 =3, 3 Ã— 1= 3.2615
3 + 3 =  6, 6 Ã— 1 = 6 and 5 + 6 =11.2630
Now we have our remainder, we can finally write down the resulting polynomial.2647
We started with x^{3} so I know this would be x^{2}.2657
x^{2} + 3x â€“ 6 and I still have my 11 being divided by x + 1.2662
Now you know pretty much everything that there is to know about dividing polynomials.2676
If you divide by a monomial, make sure you split it out among all the terms.2680
If you divide a monomial by a polynomial, you can go through the long division process 2685
or use this synthetic division process to make it nice and clean.2690
Thank you for watching www.educator.com.2694
0 answers
Post by Khanh Nguyen on October 12, 2015
Professor, there are many problems on the practice questions regarding the help given in the "show next step" button.
Could you, or someone authorized fix it? :^D
0 answers
Post by patrick guerin on July 11, 2014
Thanks for the lecture!
1 answer
Last reply by: Professor Eric Smith
Sun Jul 6, 2014 2:31 PM
Post by David Saver on July 3, 2014
You really make things easy to understand!
Thanks!!
1 answer
Last reply by: Professor Eric Smith
Tue Aug 20, 2013 2:20 PM
Post by Rana Laghaei on August 19, 2013
Thanks professor I really like your teaching style.You explain everything well and why it works.I hope you do an algebra 2 course.I appreciate your work!
0 answers
Post by Professor Eric Smith on August 12, 2013
You should put in a zero place holder any time you have a missing power in the polynomial. For example, if you are dividing by x^3 + 2x  1, then you want to put in a 0x^2 for the missing x squared term. This is the process you are seeing here in example 4. In this example the x squared term, and the single x term are both missing so we put in a place holder for each of them. Let me know if that helps out.
0 answers
Post by Ravi Sharma on August 12, 2013
How do you know when to put in zero place holders in example 4?