Factoring Polynomials - Video Tutorials & Practice Problems
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1
concept
Introduction to Factoring Polynomials
Video duration:
7m
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Hey, everyone. So in math and algebra so far, we've seen how to multiply numbers and polynomials. I could take like two times three and I know that multiplies to six and I can take this expression and do use the distributive property and know that it, it uh multiplies to X squared plus three X. Well, in some problems now, they're actually gonna give you the right side of the equation. And they're gonna ask you for the left side, they're asking you for what things you have to multiply to get here. And that's a process called factoring. And we're gonna take a look at how to do that or polynomials. And what I'm gonna show you here is that basically, whereas multiplication was taking simpler terms and multiplying into more complicated expressions. Now, we're gonna do the opposite. Factoring is the opposite of multiplication. We're gonna take a complicated expression and break it down into its simpler factors. So I'm gonna show you how to basically factor something like six into just two times three or this expression into X uh times X plus three. All right. So it turns out there's actually four ways to factor polynomials and we'll be discussing each one of them in good detail and great detail in the next couple of videos. We're gonna take a look at the first ones. I'm gonna show you how this works. The first thing you should do is you should always look for greatest common factors inside of your expressions. This has an acronym, the greatest common factor, it's called the GCF. And so I'm gonna show you how to factor out this GCF using this example over here. So we're gonna take a look at this example of two X squared plus six. And basically what the greatest common factor is is it's the largest expression that evenly divides out of each of the terms in the polynomial. I have to figure out the largest thing that divides out of everything in that polynomial. All right. And here's how to do this. Here's a step by step process. The first thing I like to do is write what's called the factor tree for each one of the terms and factor you may have seen them before. It's basically just a way to break down larger things into things that multiply. So for example, 12, there's two numbers that multiply to 12. Um I have two times six, right? You also could have used four and three and actually would have been perfectly fine. It would have worked out the same way because what happens is if I do two and six, I can't break down two anymore, but I can break down 66, actually just breaks down into two times three. So in other words, the two and the two and the three, these are all the factors of 12, I can also do the same thing for uh things with variables. In other words, the six X squared breaks down to six times X times X. But then I can also just keep breaking down the six into two times three. And I've already seen that. So in other words, the 2 to 3 and the X and the X, those are all just the factors of six X squared, we're gonna be doing this exact same thing. But now for each one of these terms in this expression, so I'm gonna do some color coding over here, I'm gonna do the two X squared and I'm gonna do the six over here. So what does the two X squared break down to? It just breaks down down into two times X times X. Can I break down anything else? No, because two can't be broken down any further. What about the six? I've seen that the six can break down into two times three. So once I've done all the factor training for each one of the terms, I'm gonna put a big parenthesis around them and uh include the sign that it was in the original expression. And now what I'm gonna do is I want to figure out the largest thing that is evenly divisible out of each one of the terms or the largest thing that pops up in between both of the terms. So if I take a look at, at this expression, what is the common items that appear in these terms? Well, I see a two that pops up in the left term and it also pops up in the right term. Is there anything else? So is the X are the X's common between both the terms? No, because they only appear on the left side. What about the three? Is that common? Well, no, because that only appears on the right term. The one thing that pops up in both is the two. So that is the greatest common factor and that leads us to now the second step, what do you do with this two? Well, basically we're gonna do is we're basically just gonna move it and extract it outside of the parentheses and kind of just remove it from this whole entire expression. And then we're gonna leave everything inside that we had from step one. So here's how this works. I take the two and I pull it to the outside of parentheses and then what was left over, what was left over was the two factors of X and then one factor of three. So in other words, what I've done here is I've done two and then I have X squared plus three. That was what was left over on the inside. And this is my greatest common factor. One easy way to check this just in case you ever worried that you didn't do it correctly is if you do the distributive property, you should basically just get back to your original expression. That's basically what the GCF is. It's like the opposite of the distributive property. All right. That's all there is to it. Let's take a look at a couple more examples here. So I have seven X squared and five X. So here what happens is I'm gonna write down the factor tree again. So I have seven X squared in five X. This just becomes seven times X times X and this just becomes five times X. Now I keep a minus sign over here and then I just put a big parenthesis. Now, can I break down the seven or five any further? Well, actually can't because seven, the only two things that multiply seven or seven and one same thing with five only things that multiply are five and one. So these are called prime factors. So what is the common items between the two? Well, it's not seven and five, but I do see one power of X that's between both of them. Why is it at two? Well, it's basically just because there's only two in the left term, but there's only one power of X in the right term. So the thing that's the largest thing that's common between both of them is just one power of X. So now what I do with this X, I pick it up and I basically just move it to the outside of the expression over here. So I read an X and then I have a parentheses and then what was left over in my expression, I have a seven X and then a five and then you always have to include the sign that was in here. So this just becomes seven X minus five. If you distribute this, you should get back to your original expression. Now, let's take a look at the last one over here. So this uh eight X, this negative eight X squared uh eight X cubed plus 16 X, all right. So let's do the factorization of this and let's do the factorization of this. Now, one of the things you actually can notice here is that eight and 16 are multiples of each other. We didn't have to worry about those two in the first two examples, but eight is just a multiple of 16. So one of the shortcuts that you can use is instead of having to factor completely. Um you can basically just write this as the product of, you can do this eight times X times X times X. And then this over here, the 16 can be written as eight times two. So notice how if I do it this way, if I notice that there are multiples of each other. I'll end up with the same thing uh in, in both of the factor trees. So now what happens is I have eight times, two times one power of X over here. Now, 11 thing that you also wanna sort of consider or be just, you know, watch out for is that I had a negative sign over here in the beginning. Uh And basically, the way I like to account for that is I like to put a little negative one times eight. So I don't forget that there's a negative sign there. All right. Now, the A plus sign over here, what are the common items between the two terms? I see an eight here and an eight over here and I see one power of X over here and one power of X over here. There's nothing else that's common. So that means that my greatest common factor is just eight X. So what do I do? I take the eight X and I pull it to the outside, all this stuff to the outside and remove it from the parentheses. And this should be just becomes eight X, this becomes eight X parentheses. And then what was left over on the inside? I see a negative 12 powers of X and then I see a two over here. So basically this is becomes negative X squared plus the two you distribute, you should get back to your original expression. All right. So these are the answers. That's how to factor using the greatest common factor. Let's go ahead and take a look at some more practice.
2
Problem
Problem
Factor out the Greatest Common Factor in the polynomial.
4x2y−100y
A
y(4x2−100)
B
4(x2y−25y)
C
4y(x2−25)
D
4y(x2−100)
3
Problem
Problem
Factor out the Greatest Common Factor in the polynomial.
−3x4+12x3−18x2
A
3x(−x2+4x−6)
B
3x2(−x2+4x−6)
C
3(−x3+4x2−6x)
D
3x2(−3x4+12x3−18x2)
4
concept
Factor by Grouping
Video duration:
4m
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Hey, everyone. So we've already seen one method of factoring polynomials, which is by factoring out a greatest common factor. So for example, if I had X cubed minus two X squared, once I wrote out the factor tree, I noticed that X squared was common in both of the terms. So I could take that and pull it to the outside. And basically, that's my factorization. I'm gonna show you this video that sometimes for some polynomials that's not going to work, you're not gonna find one common factor that works for all of the terms in this situation or this uh polynomial over here. I can't an X squared, I can't pull out a number like two or four because whatever I try to do it won't work for all the terms. I'm gonna show you in this, in this video that we're gonna need a new method called grouping to solve these types of problems. And they're very similar to greatest common factor, but there's a couple of differences. Let me just show you how it works here. First thing I want to do is actually talk about when you use these two different methods and basically just comes down to whether you can identify a common factor that works for all the terms, like for example, the X squared over here or whether you can't, and usually what's gonna happen is in these types of grouping problems, if there's gonna be four of the terms. So if you see a four term expression that you can't identify one greatest common factor for, it's usually a good indicator that's gonna be a grouping problem. And also what you'll notice is that a lot of these problems, you'll see numbers that are multiples of each other like two and four and eight and stuff like that. All right. So let me just show you how it works. Basically with the whole idea here is I'm gonna take this four term expression and I'm gonna try breaking this thing up into two different groups. That's why we call it grouping. And the hope is that I can factor out a greatest common factor from each one of the groups. So let me just show you how it works. I'm gonna show you a step by step. Process. The first thing I wanna do is make sure your polynomial is written in standard form just in case it isn't already. So in this situation, I have X cubes, two X squared, four X and eight. So this actually already is in standard form and I don't have to do anything to it. The next thing you wanna do is actually do the grouping. And basically what you're gonna do is you're gonna group terms into pairs and almost always you're just gonna do the first two in the last two terms. It's pretty much 99.9% of the time gonna work. Ok. So basically, what I like to do is I like to draw parentheses around the first two in the last two terms. And now what I'm gonna try to do is I'm gonna try to pull out a, I'm gonna try to factor out a one term, greatest common factor from each one of the groups. Basically, what I'm gonna try to do is now that I've split these things off into groups, I'm gonna turn them into problems where I just try to pull out a greatest common factor out of each one of them independently. So let's try to do this. And in fact, I actually already know what the factorization for this is. If I try to do this, it's gonna be X squared times X minus two times X squared. So I noticed that the X squared is common. I draw the little parentheses and I can pull the X squared out to the outside. This just becomes X squared. Uh And then I have X minus two. All right. So we've already seen that what happens with the second group over here? Well, for the second group, what I notice is that the four is a multiple of eight. So when I write out the factor tree, this is just four times X minus four times two. So four times two. So now it's common in this group, the common item in this group is the four. So I can take the four and move it to the outside of the expression. And what I end up with is I end up with a four here and then I end up with X minus two. So that was the third step. You're gonna factor out a one term, greatest common factor from each group. And if you notice what's happened here, we've actually ended up getting the same exact thing in both of the sort of expressions we've gotten the X minus two term. So if you think about it, now what happens is if you wrap this whole entire thing in one parenthesis, it's actually like the X minus two is now actually the common thing for both of them. So that leads us to the last step, which is now you're gonna factor out this two term, greatest common factor out of both of the groups. So it's the same thing I did with the X squared and stuff like that. I take this X minus two, pull it to the outside over here and then just write everything that's in the inside of parentheses that remains. So this just actually just becomes X minus two and then you have X squared plus four. And this actually is the complete factorization of this polynomial. If you go ahead and foil this out, which you'll actually see is that you'll end up getting back to your original expression over here. All right. All right. So again, just to summarize if you ever see a polynomial and it's four terms, and you'll notice that you can't find a common factor that for all of them, but you notice that some of the numbers are multiples of each other, try splitting up into two groups. And what you're gonna see a lot of the time is that you'll very coincidentally end up with the same factor that you can pull out of both of the groups. And so this is a very specific type of problem, but it's good to know anyway. So that's it for this one, folks. Thanks for watching.
5
Problem
Problem
Factor the polynomial by grouping
−x2−5x+7x+35
A
(−x+7)(x+5)
B
(x+7)(x+5)
C
(x−7)(x+5)
D
(x+7)(−x+5)
6
Problem
Problem
Factor the polynomial by grouping.
6x3−2x2+3x−1
A
2x2(3x−1)
B
(2x2+x)(3x−1)
C
(2x+1)(3x2−1)
D
(2x2+1)(3x−1)
7
concept
Factor Using Special Product Formulas
Video duration:
6m
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Welcome back everyone. So earlier in videos, we saw how to multiply polynomials and when multiplication problems fit a special pattern, like let's say you had X plus three and X minus three, I could use a special product formula and basically get to the right side. And so this was like X minus X squared minus nine, which is a difference of squares. In this video, we're basically gonna do the opposite. So what we're gonna see in problems is we're going to see stuff like X squared minus 36. And we have to tell whether it's going to be one of these special products here on the right side. And if we can match it to one of these formulas on the right side, then we can factor and we know that these are just going to be our factors over here. So basically, we're just doing using these special products formulas in reverse. All right. So we can also use these things to factor. That's really all there is to it. I'm gonna show you how to do this. Let's go ahead and take a look at some problems. So we have X squared minus 36. So in this case, what happens is my problem has two terms. And if you remember from your special products formulas, what happens is we have a couple of things that end up being two terms on the right side. And there's also a couple of things that end up being three terms. So all you have to do here is if this is, if this is one of our special product formulas, we're gonna look to see whether it's one of the ones that have two terms on the right side. So we're basically just trying to match this thing to one of these formulas. All right. So does this actually fit one of our special products? Well, I've got X squared minus 36 I have a perfect square which is X squared and then I have 36 which is another perfect square. These formulas over here deal with perfect with perfect cubes like a cubed and B cubes. So it's not gonna be one of these. So it turns out this whole thing is actually a squared minus B squared. That's the pattern. So that means that this factor is down to A plus B A minus B. Remember what we saw from these formulas is that these signs over here were opposite signs. So all I have to do is just figure out what my A and B are. And I'm done. So what is a? Well, if X squared is A squared that means A is X. And if 36 is B squared, that means that B is six, it's whatever I have to square to get to 36. So that means that this formula over here is just A plus B A minus B so X plus six X minus six and that's it, I'm done. And if you're ever unsure of whether you've done this correctly, you can always just multiply this expression out and you should get back to your original X squared minus 36. All right. And if you do that, you'll see that that actually does happen. So that's all there is to it. Let's take a look at the second problem here. Here, we have X cubed minus 27. Same idea. I've got two terms over here. Let's try to match into one of these equations here that have two terms on the right side. So X cubed is not a perfect square, but it's a perfect cube and 27 is not a perfect square, but it is a perfect cube. So we're not gonna use the difference of squares formula. We're actually gonna use these new equations here that we haven't yet talked about. But it's basically, you know, very similar when we have differences of cubes. So this actually turns out to be a cubed minus B cubed. And so what happens is I'm actually just gonna show you what these formulas turn out to be when you factor them you're gonna get two factors of binomial and trinomial. And I'm just going to show you how the signs work out. This is gonna be plus minus and plus, and this is going to be minus plus and plus, you'll almost never have to remember these, but just in case you ever do one way to remember it is that the last sign should always be positive, these two signs should always be opposites of each other. And the first sign should always be the sign of your expression. So here, what happens is I have an A minus B squared. So my first term is going to be A minus B same sign. And then this other term here is going to be A squared plus A B plus B squared. OK. Now, all I have to do here is just figure out what my A and B are. Well, if X cubed is equal to a cub, that means that A is X and if 27 is B cube, then B is just a number that when I multiply it by itself three times, give me 27 and it turns out that's 33 times three is 99 times three is 27. So what does this work out to? Well, this just becomes A minus B some of the words X minus three and then the second term becomes A squared. Uh So that's just X squared and then I have plus A times B. So in other words, these two things multiply together, which is three X and then B squared, B squared is just nine. So that's it. That's how this factorization happens. You would never be able to do this if you try to do this by, you know, grays, common factor or grouping or stuff like that. So these special products are really helpful. And again, if you want to multiply this out, just because you're unsure, you're gonna see that a lot of terms will cancel, but you will end up with X cubed minus 27 when you're done. OK. Last one over here we have X squared plus 12 X plus 36. So here, what I have here is I have a term or a polynomial with three terms and not two. Remember we use these equations here when we had two terms um or we try to match it to one of these that had two terms, but now I actually have three terms. So if you remember, we actually talked about some other special products called perfect square trinomial. That was these kind of equations over here where we had a squared plus two A B plus B squared, stuff like that. So which one of these is it? Well, it looks like I have two plus signs over here. So I'm gonna try to match it to one of these, the one with this one that has two plus signs. So let's see. If it actually works out, does this actually fit A squared plus two A B plus B squared? If it is, then it turns out that we can actually just factor it using this formula. If it's not, we're gonna have to use something else. OK. So does this fit? Well, if X squared is A squared, then I'm gonna guess that A is my X and if 36 is B squared, then I'm gonna guess that X, that B is my six. The tricky part in these kinds of, in looking at these types of trinomial is figuring out if this middle term actually does fit the two A B pattern. So let's find out. So this is X squared and then is this two times six or times X times six. And in fact, what actually happens is two times X times six does, in fact, gets you to add to 12 X. So basically, all it's saying is that I've identified this as being a perfect square trinomial. This is really important because if you had had something like X squared plus 10 X plus 36 now this actually doesn't fit the pattern. These are perfect squares, but this one isn't or this one doesn't fit the pattern. So you wouldn't be able to use this formula for something like that. OK. Really importantly, you try to match that. OK. So what does this become, this just becomes A plus B squared? So what's my A, my A is six. So this just becomes six X or sorry, A is X and then my B is six and this is just A X plus six squared. All right, that's it for this one, folks. Thanks for watching. Oh, last, the last point I wanted to make here um is just remember how these signs worked out. Um These were always the same signs over here. Sorry. So that's it for this one. Thanks for watching.
8
Problem
Problem
Factor the polynomial using special product formulas.
25x2−110x+121
A
(5x−10)2
B
(5x+11)(5x−11)
C
(5x+11)2
D
(5x−11)2
9
Problem
Problem
Factor the polynomial using special product formulas.
49x2−9
A
(7x−3)2
B
(7x+3)(7x−3)
C
(7x−3)2
D
(7x+3)(7x−3)
10
concept
Factor Using the AC Method When a Is 1
Video duration:
4m
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Everyone. So when we learned how to multiply polynomials, we learned how to multiply binomials by using the foil method. First, outer, inner and last. And what we saw to this multiplication is you would get X squared plus seven X plus 10. In this video, we're gonna do the reverse. We're gonna see how we can get from X squared plus seven X plus 10 or this polynomial over here and work backwards to get the two numbers that go inside the factors. Lots of people like to teach this sort of random guess and check method. But I'm gonna actually show you a systematic way to get the right answer every time. Let's go ahead and get started here. So I actually want to talk about when you use this method. And it's basically whenever you have a polynomial that fits the form A X squared plus BX plus C, anytime you have like a second degree polynomial where it's in standard form like this one over here, you're also gonna use this when you can't pull out a greatest common factor or you can't group. And most of the time what happens is you can't use special formulas. So we're gonna use this new method called the AC method. I'm gonna show you how it works. All right. So the first thing we want to do is make sure that a polynomial is in this specific form over here. And in this case, it is I have my AX squared, my BX over here, which is the five and then the, my C over here is the six. And by the way, this happens, um this is just a equals one because it's just kind of like an invisible one over here. And in fact, that's what we're going to cover in this video. We're only going to be talking about situations where this A term is equal to one. So how does this work? How do I factor out the expression over here? It's not a perfect square, it's not a formula or anything like that. Well, the first thing we're gonna do is we're gonna list out the positive and negative factors of A times C. That's why we call it the AC method. What I'm gonna do here is I like to build out this little table. So on the left side, I need to, I need numbers that multiply to A times C and then uh one times six is just six. So when A is one, this kind of just becomes only just the C term. So six. So two numbers that multiply to six and they're gonna be positive and negative So what I like to do is just start at one and then just go down from there. So one times something is six. Well, one and six is six. So what about negative one and negative six? Because you have to do positive and negative or sorry? Negative six. Well, so um let's keep going. What about two does two multiply by anything to get me to six? Yes, because two times three is six and also negative two and negative three is also six. Now, what happens if I just keep going with three? Well, if I go to 33 multiplies by two to get a six to get to six, but I already covered that pair over here. So basically what happens is um but if we keep going down here, I'm just gonna sort of get the non unique pairs. So these are just the four unique combinations that get me to sit. OK, that's the first step. The next thing we're gonna do is we're gonna find which of those factors that we just listed out add to the B term. And so let's do that. So in other words, I have to take which one of these pairs of numbers over here will add to the B and B in this case is equal to five. So let's just go and check this out. One plus six is seven. So that doesn't work negative one and negative six is negative seven. That doesn't work two and three adds to five. So that does work. But let me just be thorough. Negative two to negative three is negative five. So this doesn't work either. So once I figured out these two numbers over here, I figured out that the two numbers that add to B are two and three, your textbooks will refer to these numbers as P and Q. It's two numbers P and Q that multiply to AC, but they add to B. So this turns out to be my P and Q, it actually doesn't matter which one is, which it doesn't matter the order. All right. So basically, once you figure out these two numbers, you're done because when A is equal to one, the way that your factors are going to work out is it's going to be X plus P and X plus Q. So in other words, this expression just factors out to X plus two and X plus three. All right, or it could have been backwards. It actually doesn't matter. So if you take this expression and you foil it out, you should get back to this original expression over here. All right. That's the whole thing. So to go back to this example over here, the reason that two and five work is because two and five are two numbers that multiply to C which is 10 and they add to B which is seven. So you should always do this process of listing out the factors when you're starting out with this just to get good at it. But later on when you get quick at this, you're gonna be able to tell very quick, quickly which two numbers multiply to this number, but add to this number and you'll be able to do this pretty quickly in your head. All right. That's the AC method. Hopefully, this makes sense. Uh Thanks for watching.
11
Problem
Problem
Factor the polynomial.
x2−13x+40
A
(x+5)(x+8)
B
(x−5)(x−8)
C
(x−4)(x−10)
D
(x+4)(x+10)
12
Problem
Problem
Factor the polynomial.
x2−2x−15
A
(x−3)(x−5)
B
(x−3)(x+5)
C
(x+3)(x−5)
D
(x+3)(x+5)
13
concept
Factor Using the AC Method When a Is Not 1
Video duration:
6m
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Everyone. So we just learned how to factor a polynomial like X squared plus five X plus six. By using the AC method. First, we listed out factors of AC in our little tables over here, figured out which two numbers added to the B term. And then once we figured out those two numbers, we pop them into our binomials. And we were done, what I'm gonna show you in this video is that sometimes you might see problems that look like this in which the key difference of the polynomial is that the A term is not equal to one. So it turns out that we're going to use a lot of the same steps that we used before. But the last two steps are a little bit different in these types of situations. So I'm just gonna go ahead and show you how this works. All right. So the whole idea is that when A is not equal to one, the factoring turns out to be a little bit trickier. But we can actually bring back an old idea of grouping to solve these. Let's go ahead and just get started with this problem over here. So I have two X squared plus seven X plus six, right? So in other words, I have a polynomial that fits this form A X squared plus BX plus C in which this is my B term, my seven and this is my C term the six. So the first step is still the same. I'm gonna list my factors of A and C. What happened was when we saw these types of problems, my A was equal to just one over here and my C was six. So when, when I built out a little table, usually just all I had to do is list out the factors of six. Here, I have to list out the factors of a times six. So it's not gonna be six here. It's gonna be two times six which is 12. So that's one of the key differences here. Always make sure that you're multiplying a time. See. All right. So what are the factors of 12? Well, let's get started here. I'm gonna do one and 12 and then I have to do positive and negative. So negative one and negative 12. All right. So what, what's next? I have two is two multiply by anything to get me six. Yes, two and six does. I'm gonna do two and six. Negative two and six. What about 33 times four is 12? So three and four and then negative three and negative four. But can I keep going, do I go to four, well, four times three is 12, but I already covered that pair. So I'm done. I don't have to do it anymore. All right. See these are all the unique pairs that get me to 12. Now, I'm just gonna go ahead and add them and see which one of them, which one of those things add to the B term. That's the second step, right? So once you're done here, you just uh figure out which one of those factors add to your B term, the B term in this case is equal to seven. So I'm just gonna figure out which one of those things, which one of those pairs add to 71 in 12 gives me 13. So that's not right. This gives me negative 13. That's not right. Either two and six gives me eight. That's not right. It's close. Negative two and negative six. Give me negative eight. That's not right either. What about three and 43 and four? Give me seven. So that looks like it's right. But just to be thorough, negative three and negative four, give me negative seven. That doesn't either. So now we figured out the two numbers that multiply to 12 and add to seven before when we were dealing with this polynomial and A was equal to one, we were basically done, right? Because this was my P and this was my Q I took those numbers three and four and just popped them into my binomials. But if you try to do that here, what's gonna happen is you're gonna get two X or sorry, you're gonna get X plus three and X plus two and X, sorry, X plus four. And if you try to foil this out, what's gonna happen is you're not gonna get back to this original expression over here. In fact, you're only just gonna get X squared plus seven X plus 12. That looks nothing like this polynomial over here. So that method of just popping P and Q into these binomials is not going to work. Instead, we're going to do is that when A is not equal to one, what's going to happen is we're going to write this expression as actually a polynomial four terms. A X squared plus PXQX plus C. The whole idea here is I'm going to actually turn this polynomial into four terms. So let's do that. This becomes two X squared. And then I have A plus X again. Actually, the order doesn't really matter here. So this is three X plus four X, I could have done it backwards. It wouldn't matter. And then plus six. All right. And if you see what's kind of happened here is that the A value didn't change, the C value didn't change. But the B value I basically just broke it up into the sum of three X and four X. That's basically what happened here. So why is this helpful. Well, if you look at this or term polynomial, what you'll see here is you'll see some terms that have common factors like the two X and the four X and the three X and the six. So how do we deal with polynomials that had common factors? And usually four terms, we basically just group them. We used factoring by grouping. So I'm gonna group these things into two pairs and pull out a greatest common factor from each one of them. So this whole problem here really just turns into a grouping problem. So how do I take the two X squared and the three X and pull the greatest common factor? This is two times X times X plus three times X. So what's common with them is the X over here, I'm gonna pull that out to the outside and really, this just becomes, this becomes X and then I have two X plus three. All right. What happens with the second group? Well, this really just becomes I have four times X. In fact, actually, actually, I can break the four into two times two X plus two times three. So what's the common factor here? It's two, right? So that I'm gonna pull the two outs of this group over here and this just becomes two, what's left in the parentheses, I just have two X plus three. So it turns out that when I've done this, pulling out of the greatest common factors. I've coincidentally ended up with the same exact factor out of each group. So now what do I do? I basically just put, put a whole parentheses here and I can pull out this two X plus three all the way to the outside. And what this, he ends up becoming is two X plus three times X plus two. That's what's left inside of here. And so if you take this polynomial and there are these two factors and you foil them, you will get back to two X squared plus seven X plus six. All right. So when A is equal to one, it's a little bit more straightforward. You just pop the two numbers into your binomials and you're done when it's not equal to one, it's a little bit different. But if you follow this step here, you're actually gonna get the right answer every time again. Lots of people will teach this in sort of like a guess and check kind of way. Um But it actually gets a lot more complicated when you have stuff like two X squared and when A is not equal to one, so always use this method anyway, folks, that's it for this one. Thanks for watching.
