Algebra 1 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 complicated0010

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 us0033

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 term0054

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 2x² + 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 2x².0124

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 x², 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 term0256

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 6y².0283

That could be 1 and 6, could be 2 and 3, but they both will definitely contain y because of that y².0290

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 numbers0543

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 10q² – 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 number0788

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 = 10q², 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 10q² – 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 5t² + 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

5t² + 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 x² and y².1273

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 x² 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 x².1526

Let us take out a 2x².1529

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

14x² – 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 14x² - 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