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### 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. Don’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 – 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]

^{2}− 2x + 5 − [3/x]

[(25y

^{4}+ 60y

^{3}− 75y)/5y]

^{3}+ 12y

^{2}− 15 − [1/y]

[(54m

^{4}+ 60m

^{2}− 48m + 24)/6m]

^{3}+ 10m − 8 + [4/m]

[(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)]

^{2}+ 17p − 10)/(3p + 10)]

- [(( 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

^{3}−10x

^{2}+12x−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

^{3}+35y

^{2}−10y+65

[((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

^{3}−11a+3

*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 screen-captured 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 real-world 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) ÷ 3, I want to visualize that as the 2 ÷ 3 and also the 5 ÷ 3.*0140

*For our polynomials if I have something like (x + 3z) ÷ 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 I’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 on*0239

*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 × 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 8-6 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 × 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 I’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 × 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 × x = 2x ^{2}.*0577

*2x × 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 – 6x = 4x.*0606

*Then I have 2x ^{2} – 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 × x = 4x and 4 × 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 – 12 =0 and 4x – 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) ÷ 10m^{3}.*0777

*Since we are dividing by a monomial, I will take each of my terms and put them over what I’m dividing them by.*0784

*50m ^{4} ÷ 10m^{3}, 30m^{3} ÷10m^{3} and 20m ÷ 10m^{3}.*0792

*Now that I have done that, I will go through and simplify these one at a time.*0811

*50 ÷ 10 = 5, m ^{4} ÷ m^{3} = I can use my quotient rule and simply subtract the exponents and get m^{1}.*0816

*Continuing on 30 ÷ 10 = 3, if I subtract my exponents for m ^{3} and m^{3}, I have m^{0}.*0830

*Onto the last one, 20m ÷ 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 – anything to the 0 power is 1, 1 × 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) ÷ 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 – 2 =2 and 3 – 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 – 2 = 1 and y ^{2} – 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 – 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 × x ^{2} = 2x^{3}, that is a good sign, it is the same as the number above it.*1079

*X ^{2} × 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} – 3x^{2}, 1 – 3 = -2x^{2}.*1106

*Then I have 2x ^{3} – 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 –x.*1137

*Let us bring down some more terms.*1146

*I will get onto our multiplication process.*1149

*-x × 2x =-2x ^{2}.*1153

*-x × 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

*I’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 I’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 × 2x = 8x, 4 × 3 =12.*1228

*I can subtract this away and let us see what we get.*1238

*13 – 12=1, 8x – 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

*I’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 – 8.*1294

*All of that is being divided by x – 2.*1305

*I’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 I’m going to need x ^{2}.*1318

*Now that I found it, I will go ahead and multiply it through.*1323

*x ^{2} × x =x^{3} and x^{2} × -2 = -2x^{2}.*1327

*We have our terms, let us go ahead and subtract it away.*1336

*0x ^{2} – 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} – 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

*I’m going to need 2x.*1381

*Let us multiply that through, 2x × x =2x ^{2}.*1386

*2x × -2 =-4x.*1392

*Now we found that, let us subtract that away.*1399

*Starting on the end, 0 – -4, that is the same as 0 + 4x = 4x.*1405

*Then 2x ^{2} – 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 × x = 4x and 4 × -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} – 8) ÷ 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} – 3m^{2} – 18) ÷ 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} – 3m^{2} – 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} × 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}× 3 + 6m^{3}.*1558

*Let us subtract that away.*1568

*6m ^{3} – 6m^{3} those are gone.*1572

*I have m ^{4} – 0m^{4} I still have m^{4}.*1575

*2m ^{5} – 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} × m^{2} = m^{4}.*1609

*m ^{2} × 0m = 0m^{3} and m^{2} × 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 –m ^{2} and I’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} – 0m^{3}, 0 – 0 =0, m^{4} – 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 × 0 = 0m and -6 × 3 =-18.*1699

*It is exactly the same as the polynomial above it.*1718

*Since they are exactly the same and I’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} – 6.*1732

*In this polynomial, I have (3x ^{3} + 7x^{2} + 7x + 11) ÷ 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

*I’m going to warn you, this involves a few fractions.*1762

*Let us give it a shot.*1767

*(3x ^{3} + 7x^{2} + 7x + 11) ÷ 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} – 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 I’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 × 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 × 1/3x = 1x ^{2}.*1881

*Let us multiply that through.*1887

*1/3x × 3x = 1x ^{2} I will write it down and 1/3x × 6 = 2x.*1888

*Now we can take that and subtract it away.*1900

*7x – 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 I’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 × 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 × 3 = 5x.*1958

*5/3 × 6 =10.*1966

*We will go ahead and subtract this away.*1977

*11 – 10 = 1 and 5x – 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) ÷ 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 – number.*2037

* It will work especially with my little example right here (5x ^{3} - 6x^{2} +8) ÷ 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 I’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

*I’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 – 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, 4×5 = 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 × 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 × 50 = 200.*2205

*4 × 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 – 4.*2293

*Since it is still being divided*2312

*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} – 800) ÷ x – 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 – 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 × 6= 360.*2417

*Now we will add 0 + 360 = 360.*2430

*I have to take 360 × 6.*2435

* think I have to do a little bit of scratch work for that one.*2442

*6 × 0 = 6 × 6 = 36 and then 3 × 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 – 0= 0, 6 – 0= 6, 20 and we will breakdown and will say 11 – 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 – 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} – x^{3}) ÷ 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

*I’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 I’m going to bring down is -1 then we will go ahead and multiply.*2600

*Negative × negative is positive.*2608

*2 + 1 =3, 3 × -1= -3.*2615

*-3 + -3 = - 6, -6 × -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 – 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

1 answer

Last reply by: Professor Eric Smith

Mon Jul 31, 2017 10:29 AM

Post by Joselin ji on July 30, 2017

The pause button doesn't work, and neither does the scrubbing bar on the bottom. Does anyone know why?

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?