## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Factoring Trinomials Using the AC Method

- If the coefficient on the squared term is not one, then reverse FOIL may still work. There will be many more possibilities to check for the terms in your binomials.
- To use the AC method
- Start by making sure all terms have a greatest common factor of one
- Multiply the A and C coefficients together
- Find two numbers that multiply to give you AC but add to give you B
- Split up the middle term of the trinomial using the numbers found in step 3
- Factor the polynomial by using factor by grouping
- Be very careful with your signs when using the AC method.
- It is often helpful to write down all pairs of numbers that multiply to AC.
- When factoring is complete, check your work by using FOIL on the binomials. They should multiply to give you the original trinomial.

### Factoring Trinomials Using the AC Method

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 0:08
- Factoring Trinomials Using the AC Method 0:27
- Factoring when Leading Term has Coefficient Other Than 1
- Reversing FOIL
- Example 1 1:46
- Example 2 4:28
- Factoring Trinomials Using the AC Method Cont. 7:45
- The AC Method
- Steps to Using the AC Method
- Tips on Using the AC Method
- Example 3 10:45
- Example 4 16:50
- Example 5 21:08
- Example 6 24:58

### Algebra 1 Online Course

### Transcription: Factoring Trinomials Using the AC Method

*Welcome back to www.educator.com.*0000

*In this lesson we are going to take a look at factoring some trinomials using a method known as AC method.*0002

*We have already seen factoring trinomials once before but these ones are going to be a little bit more complicated*0010

*and that our squared term will be something other than the number 1.*0015

*This is why we are going to pick up the AC method, we have a little bit more algebraic way to approach these types of problems.*0019

*Recall some of the earlier trinomials that we have been factoring so far.*0029

*The squared term in front has always been 1 and that made life pretty easy on us*0033

*because when we went searching for those two binomials to break it down into.*0039

*We did not have a lot of options in order to get that first term.*0044

*It is probably something like y and y or x and x.*0048

*There is not a whole lot of other things it could be.*0051

*The reason why this made things a little bit easier is we only have to focus on the last term*0054

*and making sure our outside and inside terms combined to give us the middle term.*0059

*Now, we do not want to necessarily stick with those types of trinomials.*0064

*We want to go ahead and factor things where the initial term is something other than 1.*0068

*Now this will end up making things a little bit more difficult.*0073

*The good news is with these ones you can still use something like the reverse foil method.*0080

*You have not only possibilities for your first term, but now you also have possibilities for your first and last terms, both of those.*0086

*This will make it a little bit more difficult when we are checking to make sure that the outside and inside terms combine to give us that middle term.*0096

*Let us do a reverse foil example, so you can see that we have to track down many more possibilities.*0108

*This one is 2x ^{2} + 7x + 6.*0113

*It is not that big but we will look at the two binomials that we are looking to break this down into.*0117

*Like before, I will be looking for two terms that multiplied to give us the 2x ^{2}.*0124

*Since that number is there I have to look at possible things that will give us 2.*0129

*This one is not too bad, it has to be 1 and 2 to get that x ^{2}, 1x and 2x.*0136

*We will look at the possibilities that will give us our second term.*0149

* 6 could be 1 and 6, could be 2 and 3, or it could actually be those values flipped around.*0154

*The question is what should it be? What things are we looking for here?*0163

*I know that these two numbers whatever they are looks like they both better be positive since these are both positive.*0170

*It could be 1 and 6, could be 2 and 3.*0177

*Let us go ahead and put one of those in there just to see what happens.*0182

*Let us suppose I'm trying out 1and I'm trying out 6.*0187

*My first terms multiply out just fine.*0192

*My last terms multiply out just fine.*0194

*Let us check out our outside and inside terms.*0197

*The outside would give us 6x and the inside would give us 2x.*0202

*When you combine those together, you get 8x which unfortunately is not the same as that middle term.*0212

*I need to come up with some other choice for those last terms.*0219

*Let us see, if it is not 1 and 6, I guess we have to try 2 and 3.*0226

*I will put those in there and we will double check our outside and inside terms.*0232

*Outside would be 3x, inside would be 4x and sure enough those do combine to give us that 7x.*0240

*The important part to recognize is that if your initial term is something other than 1,*0250

*you have to look at your possibilities for your last term and your first term*0256

*and play around with how they are ordered in order to get your proper factorization.*0259

*Let us try another one that has a few more possibilities for that first term just to make things a little bit more interesting.*0270

*Looking at the first term, we need something that will multiply and give us 6y ^{2}.*0283

*That could be 1 and 6, could be 2 and 3, but they both will definitely contain y because of that y ^{2}.*0290

*Looking at our last terms, many different things could multiply and give us the 10.*0300

*1 and 10, 2 and 5 or possibly those just reversed around.*0307

*I have lots of different options that I can end up packaging this together.*0313

*Maybe this is 1 and 6 with the 1 and 10 or maybe I should use the 2 and 3 with the 1 and 10,*0319

*or the 1 and 6 with the 2 and 5, or the 2 and 3 with the 2 and 5.*0329

*There are many different ways and I can also do these in different orders to make sure that they combine.*0333

*The key for figuring out which combination should you use is looking at those outside and inside terms.*0340

*In this one, we want them to combine to give us that 19.*0349

*We definitely have to do a little bit of work to figure out what that is.*0352

*Let us see for the current setup my outside would be 10y and my inside would be 6y,*0355

*which unfortunately does not combine enough to give us that 19y in the middle.*0360

*I know that my 1 and 6, and my 1 and 10 I need to change something around this one.*0366

*This one is just not going to work that way.*0370

*Let us play around with our first term.*0374

*Let us try something else for the beginning here.*0378

*Let us try 2 and 3.*0381

*What will that give us?*0386

*I can see the outside is 20y, the inside is 3y but that is a little too much, 23y not going to work out.*0389

*The good news is I did do this one earlier so I do have a combination that will actually factor.*0406

*What we are looking for this one is 2, 3, 5 and 2.*0414

*Let us check the outside and inside terms for this guy.*0423

*The outside would be 4y and the inside would be 15y and sure enough those combine to give us 19y.*0427

*I know that this is the proper factorization.*0437

*Notice what this highlight is that when your initial term is something other than 1, and you have lots and lots of possibilities to run through,*0441

*it can be very difficult to find out just the right combination of numbers to use in order to make it all work out correctly.*0449

*If my numbers were even bigger I would have even more possibilities to run through.*0456

*This is a problem.*0462

*To fix this problem where our leading term is something other than 1, and our numbers could get fairly large,*0467

*we do not necessarily want to use the reverse foil method.*0474

*There are simply too many possibilities to consider for some problems and it gets too difficult.*0478

*This is why I run to pick up something known as the AC method.*0483

*This is a little bit more of an algebraic method that we can use and hunt down some of the possibilities we need for breaking it down.*0486

*Let me quickly run you through how the AC method works, and then I will give you some quick tips on using it.*0493

*The very first thing that you want to do when using AC method is just to see if everything has a common factor or not.*0499

*And if it does have a common factor, go ahead and factor that out before beginning any other type of factoring process.*0507

*If they have anything in common, pull that out.*0513

*Then I will multiply the first and the last terms together, this is known as the A and C terms.*0516

*This is where the method gets its name.*0521

*Once you get that new number, you will be looking for two numbers that multiply to give you AC and they actually add to get you B.*0524

*This will have the feel of looking for those two integers, but it will be a little bit more straightforward than what you see in the past.*0532

*I do have a nice way to organize that step to keep track of the two numbers are looking for.*0538

*Here comes the interesting part, when you find those two numbers we will actually split up your middle term into two new numbers*0543

*and then you will have four terms total and you will actually use factor by grouping to move from there.*0550

*The AC method is a way of splitting up your middle term and using a different approach that factor by grouping to handle it instead.*0556

*Let me give you some tips on using the AC method.*0565

*When you use the AC method you want to organize where your AC and your B terms go.*0572

*What I recommend is that you draw a small x.*0578

*In the top part of that x you put the values of A× C and in the bottom part of that actually put the value of B.*0585

*What you are looking to do is you want to fill out the rest of this x by putting in these two numbers.*0595

*The numbers that will go there, they must multiply to give you that top number and they must add to give you the bottom number.*0604

*You will feel like you are filling out a very small crossword.*0617

*Be very careful in doing this and make sure that the signs matchup.*0621

*If you need to add to give you a negative number then make sure the two side numbers will add to give you that negative number.*0625

*Always be careful on your signs with this one, they should matchup.*0633

*You have heard a lot about the AC method and have not seen it yet, let us go ahead and do an example see can see it all in action.*0637

*We want to factor 10q ^{2} – 23q +12.*0647

*This is a good example of one that you want use the AC method on.*0652

*If you try to factor directly you have lots of possibilities for the first term, like 1 and 10, 2 and 5.*0656

*You have lots of possibilities for the second term, 1 and 12, 2 and 6, 3 and 4.*0662

*Of course all of those reversed.*0667

*Let us tackle this using the AC method.*0671

* In step one, check all of your numbers to see if they have a common factor.*0675

*I get 10, 23 and 12 it looks like they do not have anything in common.*0679

*Unfortunately, means I can not pull anything out and make the number smaller.*0685

*In to step two, I want to go ahead and multiply my a term and my c term together.*0689

* 10 × 12 = 120.*0697

*I’m looking for two numbers that will multiply to give me 120 and add to be -23.*0702

*This is where little box will come in handy.*0709

*Let me just put in our few little notes.*0718

*We want them to multiply that gives us our top number.*0720

*We want them to add to give us that -23.*0725

*To help us better find the numbers that will go ahead and do this,*0732

*I’m going to start listing out all the pairs of numbers that will multiply to give us 120.*0736

*1 and 120, 2 and 60, I have 3 and 40, it just keep continuing making this list until you get as many numbers as possible.*0742

*I got 4 and 30, 5 and 24, 6 and 20, 8 and 15.*0757

*Now that we have a bunch of numbers on this list, let us see how we can use it.*0770

*We want these two numbers to multiply to be 120 and when we built those list that all should multiply to give us 120.*0775

*But they must add to give us -23.*0784

*The only way you want to add to get a negative number and multiply to get a positive number*0788

*is if both of these new numbers here were negative.*0792

*Let us say if you are going to pick two things off this list that will give us -23 when added together.*0797

*I think it is going to be that last two, 8 and 15.*0804

*Notice how those will multiply negative × negative will give us that positive 120 and will definitely add to be -23.*0812

*Those are the two numbers we want.*0820

*Now comes the interesting thing.*0824

*What I'm going to do with those two numbers is end up splitting up my original middle term.*0827

*I'm writing down the numbers of my original polynomial but I'm not writing that -23 in there.*0833

*This is where I split it into two terms.*0841

*This will be -8q – 15q so I have not changed my polynomial.*0845

*I just take a look at it in a different way.*0855

*You will notice how this new polynomial has four terms 1, 2, 3, 4.*0858

*I'm going to now attack it using factor by grouping, which means I will take these terms two at a time.*0863

*Let us do the first two and see what they have in common.*0871

*Both are divisible by 2q.*0875

*Let us see what we got left over in here.*0887

*2q, I have 5q - 4 and let us see what is in common with the next two.*0890

*It looks like I can pull out -3 from both of them.*0909

*That would leave us 5q - 4 and notice how the signs do match up, -3 × 5 =-15.*0923

*-3 × -4 =12.*0933

*We can go ahead and wrapped this one up.*0939

*They both have a 5q - 4 in common, I will write that for my first binomial with the leftover pieces being 2q and – 3.*0940

*It is quite a journey to get to those final two binomials, but notice how it is a little bit more methodically,*0956

*not necessarily guessing or picking things out of here.*0962

*You have a better hunting way of going about it.*0964

*Let us quickly check to make sure that this is the correct factor polynomial just by running through the foil process.*0969

*5q × 2q = 10q ^{2}, outside terms -15q, inside terms – 8q, last terms +12.*0974

*It is already starting to look pretty good since it looks like that one.*0987

*Just combine my little terms here and I get 10q ^{2} – 23q + 12, which is exactly the same as I had originally.*0991

*I know that this one is factored correctly using that AC method.*1005

*Let us see the AC method again just we can get more familiar with it.*1012

*In this one we want to factor 5t ^{2} + 13t – 6.*1016

*I want to make sure that they have a greatest common factor of only one,*1025

*which means do they have anything in common that I can factor out at the very beginning.*1028

*5, 13, and 6 do not have anything in common.*1032

*Let us move on to multiplying the A and C terms together.*1035

*5 × -6 = 30.*1041

*Two numbers that will multiply to give us 30, but add to give us a 13.*1046

*We will use our box to help us out.*1051

*It need to multiply to be 30 and add to be 13.*1057

*The two numbers that we put in here, they must multiply to give us 30 and they will add to give us 13.*1064

*To help out with the search, we will list down all the things that multiply to give us 30.*1077

*1 and 30, 2 and 15, 3 and 10, 4 and 15.*1083

*The only two things that are going to work from this list or the least that I can see I think will be our 3 and 10.*1099

*Let us go ahead and put those in.*1109

*Hold on, I think we forgot one of our signs here should be -30 on top.*1122

*We will try this again.*1131

*We need two numbers from our list that will multiply to give us -30, but add to be 13.*1133

*I think the 2 and the 15 will have to be the one to do it because they have to be different in sign to multiply to give us that -30.*1141

*How about 15 and -2?*1150

*We are ready to split up our middle term.*1156

*I have written the first term and the last term now we will write out that middle term split using these two new numbers.*1159

*– 2t + 15t looking pretty good.*1169

*Now that we have this, we want to factor by grouping.*1177

*We will take these two at a time.*1181

*5t – 2t they only have one thing in common and that would be t.*1186

*Let us see what is left over.*1199

*5t - 2 looking at the next two terms 15t – 6 they have a 3 in common.*1200

*5t – 2.*1219

*We have the 5t -2 common piece, we will go ahead and take it out of both of them and write out our leftover pieces, t + 3.*1224

*This one did take quite a bit of work let us quickly check it again by our formula.*1236

*5t ^{2} + 15t – 2t -6.*1242

*These two little terms combine giving us 13t - 6, which is the same as the original.*1251

*I know that this is the correct factorization.*1260

*Let us go ahead and try to factor this one using the AC method.*1269

*This one has a few more variables in it so notice I have x ^{2} and y^{2}.*1273

*Do not worry too much about those x and y, what you will see just focus mainly on those numbers and hunt down what those need to be in.*1278

*Is there anything I can take out at the very beginning?*1288

*Do they have anything in common?*1291

*It looks like the 11 is going to make it where they do not have anything in common.*1295

*I will multiply the A and C terms together.*1301

*6 × -10 = -60.*1304

*We need two numbers that will multiply to be -60, but add to be 11.*1312

*Let us try our box to help us out.*1317

*-60 and 11.*1322

*They need to multiply and give us -60 and add to be the 11.*1327

*Starting off with writing down all the possibilities to give is that 60.*1337

*1 and 60 would do it, 2 and 30, 3 and 20, 4 and 15.*1341

*The two numbers that we use must multiply to be -60.*1353

*We want one of these numbers to be positive and the other one to be negative.*1357

*Since we are adding to be 11, the larger number must be positive.*1362

*I think I see a good option on this list, the 4 and the 15.*1368

*The 15 is the larger one, so it must be positive and the 4 is the smaller one, so we will make it negative.*1372

*We will write down our polynomial and split up the middle term into two new terms.*1380

*We will use 15xy and - 4xy.*1391

*Notice that we are using that xy here because those are what is on the middle term.*1398

*Onto that factor by grouping process.*1405

*We will look at our first two terms and we will look at our second two terms.*1408

*Starting with the first two, what do they have in common?*1415

*They both have a 3x but I can go ahead and draw out.*1418

*That would leave me a 2x + 5y.*1426

*Looking at the next two terms, they have a -2y in common.*1434

*What would that leave?*1446

*2x + 5y.*1449

*Double check and make sure your signs will get -2 × 2 =-4, -2 × 5 =-10.*1454

*Things are looking good.*1461

*Okay, now I have that common piece we will take it out from both of them.*1464

*2x + 5y and then 3x -2y.*1472

*It looks like this one is factored and you can double check by running through the foil process.*1482

*After all we get a little bit more familiar with that AC method in seeing the steps as you go along.*1488

*Keep your components, find those two new numbers so you can split up that middle term.*1492

*Let us do one last example and I want to do one where they had a greatest common factor that you could pull out.*1500

*You will notice how initially these numbers are quite large, we have 28, 58 and -30.*1508

*Furthermore, they all have an x ^{2} in common.*1514

*Before we even start some of the AC method, let us go ahead and pull out that common term first.*1516

*What do they have in common?*1523

*Everything is divisible by 2 and they all have an x ^{2}.*1526

*Let us take out a 2x ^{2}.*1529

*Let us see if this makes things a little bit smaller.*1534

*14x ^{2} – 58 ÷ 2 = 29x and this one divided by 2 – 15.*1537

*It looks like they do not have anything else in common.*1557

*We need to continue factoring it.*1559

*From here I’m going to take its A and C term and I’m going to multiply those together.*1562

*Now these ones are quite large, but we can do it.*1569

*14 × -15 = -210 quite large.*1573

*I need two numbers that multiply to give me -210, but they add to be -29.*1581

*Let us draw our box and start hunting down some possibilities.*1589

*I have -210, must add to be -29.*1594

*I want to make this easy on ourselves, reduce it as easy as possible.*1602

*I’m writing down possibilities that will multiply to be 210.*1605

*1 and 210, 2 and 105, 3 and 70, 5 and 42, 6 and 35, 7 and 30, 10 and 21, 14 and 50.*1609

*Now that I have a list of bunch of different numbers, we need to multiply to give us -210.*1633

*That means one of these numbers will be negative and one of them will be positive.*1640

*They will add to be -29 so I know the larger number must be a negative.*1647

*It is the only way we will get -29 when adding.*1653

*Come over the list very carefully, the one that will do it will be this pair right here, the 6 and 35.*1657

*Sure enough, those multiply to give us the -210 and they add to give us -.29.*1672

*Sometimes you might have to go through and check in these one by one, but it is worthy process.*1678

*We have that and we are going to write down the 14x ^{2} - 15.*1685

*Let us take our middle term and split it up 6x - 35x.*1693

*Right now, we can continue with our factor by grouping.*1703

*Grabbing this first two and looking at what they have in common.*1708

*I see that we can take out 2x.*1712

*That would leave us with 7x + 3.*1719

*Looking at the next two numbers.*1729

*These ones we can pull out -5.*1732

*That would leave us with 7x and 3 looking pretty good.*1742

*Now I can grab my two binomials and almost be done.*1750

*They have a 7x + 3 in common.*1756

*There would be 2x - 5 left over.*1759

*Be careful this one is not done.*1763

*Remember that initial factor we took out at the very beginning, it is still out front of this entire process.*1766

*Feel free to write it down now in front of all of this.*1774

*That way you do not forget it.*1778

*Now we have the final factored form of our polynomial.*1781

*Using the guess and check method and the AC method can be two great ways to factor your polynomials.*1788

*I suggest using the reverse foil method if the numbers are not that big or the coefficient is 1.*1794

*It starts to get a little bit more complicated then feel free use this AC method to break it down.*1801

*Thank you for watching www.educator.com.*1807

1 answer

Last reply by: Professor Eric Smith

Mon Jan 26, 2015 8:49 PM

Post by julius mogyorossy on January 26, 2015

That is very cool, how did somebody think of it.

1 answer

Last reply by: Professor Eric Smith

Thu Oct 10, 2013 1:57 PM

Post by Sophia Zimmer on October 9, 2013

We tried to use the AC method to factor the following trinomials, but we keep getting three different terms no matter what we do.

14x(2)-53x+14

99q(2)-92q+9

()indicates powers

Is there another way of looking at these trinomials that may help us solve them?

Thanks! :-)

0 answers

Post by Professor Eric Smith on October 3, 2013

5 and 12 are two great numbers that do multiply to give us 60. Remember we also want them to add and give us 11. No matter which number has the negative sign we can't make them add to 11. -5 + 12 = 7 or 5 - 12 = -7

For this reason, we do not use them to split our middle term.

0 answers

Post by Veronica Perez on September 22, 2013

Why wouldn't 5 and 12 be two numbers that add up to -60 on example 5, if you don't mind me asking?