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.