Factoring Polynomials - Online Tutor, Practice Problems & Exam Prep

Factoring polynomials involves breaking down complex expressions into simpler factors. Key methods include identifying the greatest common factor (GCF) and using grouping for four-term polynomials. Special product formulas, such as the difference of squares and perfect square trinomials, aid in recognizing patterns for efficient factoring. The AC method is essential when the leading coefficient is not one, requiring the identification of factors that multiply to ac and add to b. Mastering these techniques enhances problem-solving skills in algebra.

Introduction to Factoring Polynomials

Hey, everyone. So in math and algebra, so far, we've seen how to multiply numbers and polynomials. I could take, like, 2 times 3, and I know that multiplies to 6. And I can take this expression and do use the distributive property and know that it multiplies to x2 + 3x. Well, in some problems now, they're actually going to give you the right side of the equation, and they're going to 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 going to take a look at how to do that for polynomials. And what I'm going to show you here is that, basically, whereas multiplication was taking simpler terms and multiplying into more complicated expressions, now we're going to do the opposite. Factoring is the opposite of multiplication. We're going to take a complicated expression and break it down into its simpler factors. So I'm going to show you how to basically factor something like 6 into just 2 times 3 or this expression into x×x+3. Alright?

So it turns out there are actually 4 ways to factor polynomials, and we'll be discussing each one of them in great detail in the next couple of videos. We're going to take a look at the first ones. I'm going to show you how this works. The first thing you should do is you should always look for greatest common factors inside your expressions. This has an acronym, the greatest common factor. It's called the GCF, and so I'm going to show you how to factor out this GCF using this example over here. So we're going to take a look at this example of 2x2+6 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. Alright?

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 a factor tree for each one of the terms. In factor trees, 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. I have 2 times 6. Right? I could also have used 4×3 and would have been perfectly fine. So if I do 2 times 6, I can't break down 2 anymore, but I can break down 6, which actually just breaks down into 2 times 3. So in other words, the 2 and the 2 and the 3, these are all the factors of 12. I can also do the same thing for things with variables. In other words, the 6x2 breaks down to 6 times x times x, but then I can also just keep breaking down the 6 into 2 times 3, and I've already seen that.

So in other words, the 2, the 3, and the x and the x, those are all just the factors of 6x2. We're going to be doing this exact same thing, but now for each one of these terms in this expression. I'm going to do some color coding over here. I'm going to do the 2x2, and I'm going to do the 6 over here. So what does the 2x2 break down to? It just breaks down into 2 times x times x. Can I break down anything else? No. Because 2 can't be broken down any further. What about the 6? I've seen that the 6 can break down into 2 times 3.

So once I've done all the factor training for each one of the terms, I'm going to put a big parenthesis around them and include the sign that it was in the original expression. And now what I'm going to 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 this expression, what are the common items that appear in these terms? Well, I see a 2 that pops up in the left term, and it also pops up in the right term.

Are the x's common between both the terms? No. Because they only appear on the left side. What about the 3? Is that common? Well, no. Because that only appears on the right term. The one thing that pops up in both is the 2, so that is the greatest common factor. That leads us to now the second step. What do you do with this 2? Well, basically, what we're going to do is we're basically just going to move it and extract it outside of the parentheses and kind of just remove it from this whole entire expression, and then we're going to leave everything inside that we had from step 1. So here's how this works. I take the 2 and I pull it to the outside of parentheses. And then what was left over?

What was left over were the 2 factors of x and then one factor of 3. So, in other words, what I've done here is I've done 2, and then I have x2+3. That was what was left over on the inside, and this was my greatest common factor. One easy way to check this, just in case you ever worry 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. Alright? That's all there is to it. Let's take a look at a couple more examples here.

So I have 7x2 and 5x. Here what happens is I'm going to write down the factor tree again. So I have 7x2 and 5x. This just becomes 7 times x times x, and this just becomes 5 times x. Now I keep a minus sign over here, and then I just put a big parenthesis.

Can I break down the 7 or 5 any further? Well, actually, I can't because 7, the only two things that multiply 7 are 7 and 1. Same thing with 5. Only things that multiply are 5 and 1. So these are all prime factors. What are the common items between the two? Well, it's not 7 and 5, but I do see one power of x that's between both of them. Why is it x? Well, it's basically just because there's only x 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 do 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 write an x, and then I have a parenthesis. And then what was left over in my expression? I have a 7x and then a 5, and then you always have to include the sign that was in here. So this just becomes 7x-5. 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, 8x this negative 8x3 plus 16x. Alright. 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 8 and 16 are multiples of each other. We didn't have to worry about those two in the first two examples, but 8 is just a multiple of 16. So one of the shortcuts that you can use is instead of having to factor completely,

Factor out the Greatest Common Factor in the polynomial.

$4x^2y-100y$

• A. $y\left(4x^2-100\right)$
• B. $4\left(x^2y-25y\right)$
• C. $4y$($x^2-25$)
• D. $4y\left(x^2-100\right)$

Factor out the Greatest Common Factor in the polynomial.

$-3x^4+12x^3-18x^2$

• A. $3x\left(-x^2+4x-6\right)$
• B. $3x^2\left(-x^2+4x-6\right)$
• C. $3\left(-x^3+4x^2-6x\right)$
• D. $3x^2\left(-3x^4+12x^3-18x^2\right)$

Factor by Grouping

Everyone. So we've already seen one method of factoring polynomials, which is by factoring out the greatest common factor. For example, if I had x³ minus 2x², once I wrote out the factor tree, I noticed that x² 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 going to show you in this video that sometimes for some polynomials, that's not going to work. You're not going to find one common factor that works for all of the terms. In this situation or this polynomial over here, I can't pull out an x². I can't pull out a number like 2 or 4 because whatever I try to do, it won't work for all the terms. I'm going to show you in this video that we're going to need a new method called grouping to solve these types of problems. They're very similar to the greatest common factor, but there are a couple of differences. Let me just show you how it works here. The first thing I want to do is actually talk about when you use these two different methods and, basically, it just comes down to whether you can identify a common factor that works for all the terms, like, for example, the x² over here, or whether you can't. And, usually, what's going to happen in these types of grouping problems is there's going to be 4 of the terms. So if you see a 4 term expression that you can't identify one greatest common factor for, it's usually a good indicator that it's going to be a grouping problem. Also, what you'll notice is that many of these problems, you'll see numbers that are multiples of each other, like 24 and 8 and such.

Alright? So let me just show you how it works. Basically, the whole idea here is I'm going to take this 4 term expression, and I'm going to try breaking this thing up into 2 different groups. That's why we call it grouping. And the hope is that I can factor out the greatest common factor from each one of the groups. So let me just show you how it works. I'm going to show you a step-by-step process. The first thing you want to do is make sure your polynomial is written in standard form just in case it isn't already. So in this situation, I have x³, 2x², 4x, and 8. So this actually already is in standard form, and I don't have to do anything to it. The next thing you want to do is actually do the grouping. And, basically, what you're going to do is you're going to group terms into pairs, and almost always, you're just going to do the first two and the last two terms. It's pretty much 99.9% of the time going to work. Okay. So, basically, what I like to do is I like to drop parentheses around the first two and the last two terms. And now what I'm going to try to do is I'm going to try to factor out a one term greatest common factor from each one of the groups. Basically, what I'm going to try to do is now that I've split these things off into groups, I'm going to 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. In fact, I actually already know what the factorization for this expression is. If I try to do this, it's going to be x² times (x - 2) for the first group. So I notice that the x² is common. I draw the little parentheses, and I can pull the x² out to the outside. This just becomes x² x - 2. Alright. 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 4 is a multiple of 8. So when I write out the factor tree, this is just 4 times (x - 2). So 4 times 2. So now what's common in this group? The common item in this group is the 4. So I can take the 4 and move it to the outside of the expression, and what I end up with is I end up with a 4 here and then I end up with x - 2. So that was the third step. You're going to factor out a one term greatest common factor from each group. And if you'll 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 - 2 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 - 2 is now actually the common thing for both of them. So that leads us to the last step, which is now you're going to factor out this 2 term greatest common factor out of both of the groups. So it's the same thing I did with the x² and such. I take this x - 2, and pull it to the outside over here, and then just write everything that's inside of parentheses that remains. So this just actually just becomes x - 2, and then you have x² + 4. And this actually is the complete factorization of this polynomial. If you go ahead and foil this out, what you'll actually see is that you'll end up getting back to your original expression over here. Alright? Alright. So, again, just to summarize, if you ever see a polynomial and it's 4 terms and you notice that you can't find a common factor that works for all of them, but you notice that some of the numbers are multiples of each other, try splitting up into 2 groups. And what you're going to 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.

Factor the polynomial by grouping

$-x^2-5x+7x+35$

• A. $\left(-x+7\right)\left(x+5\right)$
• B. $\left(x+7\right)\left(x+5\right)$
• C. $\left(x-7\right)\left(x+5\right)$
• D. $\left(x+7\right)\left(-x+5\right)$

Factor the polynomial by grouping.

$6x^3-2x^2+3x-1$

• A. $2x^2\left(3x-1\right)$
• B. $\left(2x^2+x\right)\left(3x-1\right)$
• C. $\left(2x+1\right)\left(3x^2-1\right)$
• D. $\left(2x^2+1\right)\left(3x-1\right)$

Factor Using Special Product Formulas

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, I had x + 3 and x - 3, I could use a special product formula and basically get to the right side. And so this was like x^2 - 9, which is a difference of squares. In this video, we're basically going to do the opposite. So what we're going to see in problems is we're going to see stuff like x^2 - 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 using these special product formulas in reverse. Alright? So we can also use these things to factor. That's really all there is to it. I'm going to show you how to do this. Let's go ahead and take a look at some problems.

So we have x^2 - 36. So in this case, what happens is my problem has 2 terms. And if you remember from your special products formulas, what happens is we have a couple of things that end up being 2 terms on the right side. There's also a couple of things that end up being 3 terms. So all you have to do here is if this is one of our special product formulas, we're going to 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. Alright? So does this actually fit one of our special products? Well, I've got x^2 - 36. I have a perfect square, which is x^2, and then I have 36, which is another perfect square. These formulas over here deal with perfect cubes, like a^3 and b^3, so it's not going to be one of these. So it turns out this whole thing is actually a^2 - b^2. That's the pattern. So that means that this factors down to a + b, a - 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^2 is a^2, that means a is x. And if 36 is b^2, that means that b is 6. It's whatever I have to square to get to 36. So that means that this formula over here is just a + b, a - b. So x + 6, x - 6, 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^2 - 36. Alright? 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^3 - 27. Same idea. I've got 2 terms over here. Let's try to match it to one of these equations here that have two terms on the right side. So x^3 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 going to use the difference of squares formula. We're actually going to 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^3 - b^3. And so what happens is I'm actually just going to show you what these formulas turn out to be. When you factor them, you're going to get 2 factors of binomial and trinomial. And I'm just going to show you how the signs work out. This is going to 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 original expression. So here what happens is I have a^3 - b^3, So my first term is going to be a - b. Same sign. And then this other term here is going to be a^2 + ab + b^2. Okay? Now all I have to do here is just figure out what my a and b are. Well, if x^3 is equal to a^3, that means that a is x. And if 27 is b^3, then b is just a number that when I multiply it by itself 3 times gives me 27. And it turns out that's 3. 3 times 3 is 9, 9 times 3 is 27. So what does this work out to? Well, this just becomes a - b, in other words, x - 3. And then the second term becomes a^2, so that's just x^2. And then I have plus a times b, so in other words, these two things multiply together, which is 3x. And then b^2, b^2 is just 9. 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, greatest 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 going to see that a lot of terms will cancel, but you will end up with x^3 - 27 when you're done. Okay? Last one over here, we have x^2 + 12x + 36. So here, what I have here is I have a term or a polynomial with 3 terms and not 2. Remember, we use these equations here when we had 2 terms, or we try to match it to one of these that had 2 terms, but now I actually have 3 terms. So if you remember, we actually talked about some other special products called perfect square trinomials. That was these kinds of equations over here where you had a^2 + 3b + b^2, stuff like that. So which one of these is it? Well, it looks like I have 2 plus signs over here, so I'm going to try to match it to one of these. The one this one that has 2 plus signs. So let's see if it actually works out. Does this actually fit a^2 + 2ab + b^2? If it is, then it turns out that we can actually just factor using this formula. If it's not, we're going to have to use something else.

Okay? So does this fit? Well, if x^2 is a^2, then I'm going to guess that a is my x. And if 36 is b^2, then I'm going to guess that that b is my 6. The tricky part in looking at these types of trinomials is figuring out if this middle term actually does fit the 2ab pattern. So let's find out. So this is x^2, and then is this 2 \times 6 \times x? And in fact, what actually happens is 2 \times x \times 6 does, in fact, get you to 12x. 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^2 + 10x + 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. Okay? Really importantly, you try to match that. Okay. So what does this become? This just becomes (x + 6)^2. Alright? That's it for this one, folks. Thanks for watching. Oh, last last point I wanted to make here, is just remember how these signs worked out. These were always the same signs over here. Sorry. So that's it for this one. Thanks for watching.

Factor the polynomial using special product formulas.

$25x^2-110x+121$

• A. $\left(5x-10\right)^2$
• B. $\left(5x+11\right)\left(5x-11\right)$
• C. $\left(5x+11\right)^2$
• D. $\left(5x-11\right)^2$

Factor the polynomial using special product formulas.

$\frac{x^2}{49}-9$

• A. $\left(\frac{x}{7}-3\right)^2$
• B. $\left(\frac{x}{7}+3\right)\left(\frac{x}{7}-3\right)$
• C. $\left(7x-3\right)^2$
• D. $\left(7x+3\right)\left(7x-3\right)$

Factor Using the AC Method When a Is 1

Hey, 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 out of this multiplication is you would get \( x^2 + 7x + 10 \). In this video, we're going to do the reverse. We're going to see how we can get from \( x^2 + 7x + 10 \) 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 going to 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 \( ax^2+bx+c \) anytime you have a second degree polynomial where it's in standard form, like this one over here. You're also going to 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 going to use this new method called the AC method. I'm going to show you how it works. Alright?

So the first thing we want to do is make sure that our polynomial is in this specific form over here, and in this case, it is. I have my \( a \, x^2 \), my \( b \, x \) over here, which is the 5, and then the my \( c \) over here is the 6. And by the way, this happens, this is just \( a = 1 \) because it's just kind of like an invisible one over here. 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 1.

So how does this work? How do I factor out this expression over here? It's not a perfect square. It's not a formula or anything like that. Well, the first thing we're going to do is we're going to list out the positive and negative factors of \( a \times c \). That's why we call it the AC method. What I'm going to do here is I like to build out this little table. So on the left side, I need two numbers that multiply to \( a \times c \), and then, one times 6 is just 6. So when \( a \) is 1, this kind of just becomes only just the \( c \) term. So 6. So two numbers that multiply to 6, and they're going to be positive and negative. So what I like to do is just start at 1 and then just go down from there. So one times something is 6, well, 1x6 is 6. So what about negative one and negative six? Because you have to do positive and negative, or sorry, negative 6. Well, so, let's keep going. What about 2? Does 2 multiply by anything to get me to 6? Yes. Because 2 times 3 is 6, and also negative 2 and negative 3 is also 6. Now, what happens if I just keep going with 3? Well, if I go to 3, 3 multiplies by 2 to get to 6, but I already covered that pair over here. So basically what happens is, but if we keep going down here, I'm just going to sort of get the non-unique pairs. So these are just the 4 unique combinations that get me to 6. Okay? So that's the first step. The next thing we're going to do is we're going to 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 5. So let's just go and check this out. 1+6 is 7, so that doesn't work. Negative one and negative six is negative seven, that doesn't work. 2+3 adds to 5, 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 2 and 3. Your textbooks will refer to these numbers as \( p \) and \( q \). It's 2 numbers \( p \) and \( q \) that multiply to \( a \times c \), 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.

So basically, once you figure out these two numbers, you're done because when \( a \) is equal to 1, the way that your factors are going to work out is it's going to be \( x + p \) and \( x + q \). So in other words, this expression just factors out to \( x + 2 \) and \( x + 3 \). Alright? 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. Alright? That's the whole thing. So to go back to this example over here, the reason that 2 and 5 work is because 2 and 5 are two numbers that multiply to \( c \), which is 10, and they add to \( b \), which is 7. 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 going to be able to tell very 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. Alright? That's the AC method. Hopefully, this makes sense. Thanks for watching.

Factor the polynomial.

$x^2-13x+40$

• A. $\left(x+5\right)\left(x+8\right)$
• B. $\left(x-5\right)\left(x-8\right)$
• C. $\left(x-4\right)\left(x-10\right)$
• D. $\left(x+4\right)\left(x+10\right)$

Factor the polynomial.

$x^2-2x-15$

• A. $\left(x-3\right)\left(x-5\right)$
• B. $\left(x-3\right)\left(x+5\right)$
• C. $\left(x+3\right)\left(x-5\right)$
• D. $\left(x+3\right)\left(x+5\right)$

Factor Using the AC Method When a Is Not 1

Hey, everyone. So we just learned how to factor a polynomial like \(x^2 + 5x + 6\) 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 this polynomial is that the \(a\) term is not equal to 1. So it turns out that we're gonna 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. Alright. So the whole idea is that when \(a\) is not equal to 1, 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 \(2x^2 + 7x + 6\). Alright? So in other words, I have a polynomial that fits this form \(ax^2 + bx + c\), in which this is my \(b\) term, my 7, and this is my \(c\) term, the 6. 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 1 over here and my \(c\) was 6. So when I built out a little table, usually, all I had to do is list out the factors of 6. Here, I have to list out the factors of \(a \times 6\). So it's not gonna be 6 here. It's gonna be \(2 \times 6\), which is 12. So that's one of the key differences here. Always make sure that you're multiplying \(a \times c\). Alright? So what are the factors of 12? Well, let's get started here. I'm gonna do \(1 \times 12\), and then I have to do positive and negative. So negative one and negative twelve. Alright. So what's next? I have 2. Does 2 multiply by anything to get me 6? Yes, \(2 \times 6\) does. I'm gonna do 2 and 6, negative 2 and 6. What about 3? \(3 \times 4\) is 12, so 3 and 4, and then negative 3 and negative 4. But can I keep going? Do I go to 4? Well, \(4 \times 3\) is 12, but I already covered that pair. So, I'm done. I don't have to do it anymore. Alright? 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 adds to the \(b\) term. That's the 2nd step. Right? So once you're done here, you just figure out which one of those factors add to your \(b\) term. The \(b\) term, in this case, is equal to 7, so we're just gonna figure out which one of those things, which of those pairs add to 7. \(1 \times 12\) gives me 13, so that's not right. This gives me negative 13, that's not right either. \(2 \times 6\) gives me 8. That's not right. It's close. So negative 2 negative 6 give me negative 8. That's not right either. What about \(3 \times 4\)? \(3 \times 4\) gives me 7, so that looks like it's right. But just to be thorough, negative three and negative four give me negative seven, that doesn't work either. So now, we figure out the two numbers that multiply to 12 and add to 7. Before, when we were dealing with this polynomial and \(a\) was equal to 1, we were basically done. Right? Because this was my \(p\) and this was my \(q\). I took those numbers \(3 \times 4\) and just popped them into my binomials. But if you try to do that here, what's gonna happen is you're gonna get \(x + 3\) and \(x + 4\). 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 \(x^2 + 7x + 12\). 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, what we're gonna do is that when \(a\) is not equal to 1, what's gonna happen is we're gonna write this expression as actually a polynomial of 4 terms. \(Ax^2+px+qx+c\). The whole idea here is I'm gonna actually turn this polynomial into 4 terms. So let's do that. This becomes \(2x^2\), and then I have, plus \(px\). Again, it actually the order doesn't really matter here. So this is \(3x + 4x\). I could have done it backward, it wouldn't matter, and then plus 6. Alright? 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 \(3x\) and \(4x\). That's basically what happened here. So why is this helpful? Well, if you look at this 4 term polynomial, what you'll see here is you'll see some terms that have common factors like the \(2x\) and the \(4x\) and the \(3x\) and the \(6\). 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 2 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 \(2x^2\) and the \(3x\) and pull the greatest common factor? This is \(2x \times x\), plus \(3x\). So, what's common with them is the \(x\) over here. I'm going to pull that out to the outside and really this just becomes \(x\), and then I have \(2x + 3\). Alright? What happens with the second group? Well, this really just becomes I have \(4x\). In fact, actually, I can break the 4 into \(2x \times 2x + 2x \times 3\). So what's the common factor here? It's 2. Right? So then I'm gonna pull the 2 out of this group over here, and this just becomes 2. What's left in the parenthesis? I just have \(2x+3\). 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 a whole parenthesis here, and I can pull out this \(2x + 3\) all the way to the outside. And what this ends up becoming is \(x(2x + 3) + 2(2x + 3)\), which simplifies further to \((2x + 3)(x + 2)\). And so if you take this polynomial and it were these two factors and you foil them, you will get back to \(2x^2 + 7x + 6\). Alright? So when \(a\) is equal to 1, 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 1, 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, but it actually gets a lot more complicated when you have stuff like \(2x^2\) and when \(a\) is not equal to 1. So always use this method. Anyway, folks, that's it for this one. Thanks for watching.

Factor the polynomial.

$4x^2-19x+12$

• A. $\left(4x-3\right)\left(x-4\right)$
• B. $\left(4x-6\right)\left(x-2\right)$
• C. $\left(2x-3\right)\left(2x-4\right)$
• D. $\left(2x-6\right)\left(2x-2\right)$

Factor the polynomial.

$3x^2-2x-5$

• A. $\left(x+3\right)\left(x-5\right)$
• B. $\left(x+1\right)\left(x-5\right)$
• C. $\left(3x+1\right)\left(x-5\right)$
• D. $\left(x+1\right)\left(3x-5\right)$

Here’s what students ask on this topic:

What is the greatest common factor (GCF) in polynomial factoring?

The greatest common factor (GCF) in polynomial factoring is the largest expression that evenly divides all terms in the polynomial. To find the GCF, you break down each term into its prime factors and identify the common factors. For example, in the polynomial ${\mathrm{2x}}^2+\; 6$, the GCF is 2 because it is the largest number that divides both terms. Factoring out the GCF simplifies the polynomial, making it easier to work with. The process involves writing the polynomial as a product of the GCF and the remaining terms.

How do you factor polynomials by grouping?

Factoring by grouping is used when a polynomial has four terms. The process involves grouping the terms into pairs and factoring out the GCF from each pair. For example, consider the polynomial $x^3-\; 2x^2+\; 4x\; -\; 8$. First, group the terms: $(x^3-\; 2x^2)\; +\; (4x\; -\; 8)$. Next, factor out the GCF from each group: $x{x}^2(x\; -\; 2)\; +\; 4(x\; -\; 2)$. Finally, factor out the common binomial factor: $(x\; -\; 2)(x^2+\; 4)$.

What are special product formulas in polynomial factoring?

Special product formulas in polynomial factoring include patterns like the difference of squares and perfect square trinomials. The difference of squares formula is $a^2-b^2=\; (a\; +\; b)(a\; -\; b)$. For example, $x^2-\; 36$ factors to $(x\; +\; 6)(x\; -\; 6)$. Perfect square trinomials follow the pattern $a^2+\; 2ab\; +b^2=\; (a\; +\; b){}^{}$. For example, $x^2+\; 12x\; +\; 36$ factors to $(x\; +\; 6){}^{}$.

How does the AC method work for factoring polynomials?

The AC method is used for factoring polynomials of the form $ax{}^{}+\; bx\; +\; c$ when the leading coefficient $a$ is not 1. First, multiply $a$ and $c$. List the factors of $ac$ and find the pair that adds to $b$. Rewrite the middle term using these factors, then group and factor by grouping. For example, for $2x{}^{}+\; 7x\; +\; 6$, multiply $2$ and $6$ to get $12$. The factors $3$ and $4$ add to $7$. Rewrite as $2x{}^{}+\; 3x\; +\; 4x\; +\; 6$, then group: $x(2x\; +\; 3)\; +\; 2(2x\; +\; 3)$. Factor out the common binomial: $(2x\; +\; 3)(x\; +\; 2)$.

What is the difference between factoring by grouping and the AC method?

Factoring by grouping is used for polynomials with four terms, where you group terms into pairs and factor out the GCF from each pair. For example, $x^3-\; 2x^2+\; 4x\; -\; 8$ is grouped as $(x^3-\; 2x^2)\; +\; (4x\; -\; 8)$, then factored to $x{}^{}+\; 4(x\; -\; 2)$, and finally $(x\; -\; 2)(x{}^{}+\; 4)$. The AC method is used for trinomials of the form $ax{}^{}+\; bx\; +\; c$ when $a$ is not 1. It involves multiplying $a$ and $c$, finding factors that add to $b$, rewriting the polynomial, and then grouping. For example, $2x{}^{}+\; 7x\; +\; 6$ is rewritten as $2x{}^{}+\; 3x\; +\; 4x\; +\; 6$, grouped as $x(2x\; +\; 3)\; +\; 2(2x\; +\; 3)$, and factored to $(2x\; +\; 3)(x\; +\; 2)$.