 In this example, we're going to write a partial fraction decomposition for the rational function x over x squared minus 5x plus 6. So the first thing to do when you do a partial fraction decomposition is to check, is it a proper fraction? Which in this case, the answer is yes. The numerator is degree 1, the denominator is degree 2. So this is a proper fraction. If it was improper, we'd have to do long division. So we go to the next step. The next step here is your need to factor the denominator. So we have this denominator x squared minus 5x plus 6. It was a horrible, horrible thing that someone multiplied it out. It's like you had your perfect figurine of all of your bobblehead Marvel characters, and then the three-year-olds came and knocked them all over. It's a horrible thing to do, multiplying out a denominator. Now this one's not too hard to pick up. We need factors of 6 that add up to be negative 5. We can get away with x minus 3 and x minus 2 as our magic pair. And so then that then gives us our denominator, right? So we can rewrite the original fraction as x over x minus 3 times x minus 2. And this process is only as difficult as factoring the denominator can be. And so for the most part, you're going to see examples in the corresponding homework that the deniers are typically factored, because we don't want this to be a huge factoring problem, but you would have to factor them at this step. The next part, oftentimes you can skip one, steps one and two, because the question is generous in that regard. The next step is to look for the template. Back in the day, Microsoft Word had a good friend, which we called Clippy, right? He was a paperclip for which if you started typing in things like to whom this may concern or start typing your resume, Clippy would appear on the screen and be like, I see that you're writing a letter. I see that you're writing a resume or something like that. And they give you hints on how you can do this. Like, oh, I have some templates already made for you. A template meaning that the document is started for you, but it has some fields that you have to fill in your own personal information, because Clippy doesn't know what your last job was or anything like that. So we wanted the same thing with our partial fraction decomposition. Our template will look something like the following, our original fraction x times x minus 3 and x minus 2. Because after all, what we're trying to do is we're trying to come up with two partial fractions, two fractions would add together to give us this fraction for which when we look at the denominator, the denominator is supposed to be the least common denominator of these two fractions right here. So this actually tells us what the denominators are going to be. The denominators are going to be x minus 3 and x minus 2. Because if we add these two fractions together, the least common denominator would be x minus 3 times x minus 2. And therefore, let me highlight that this is a 3 right here. And so the denominators have to combine to give x minus 3 and x minus 2. So our partial fractions are going to look like this. What are the numerators? Well, this is where the idea about to be a proper fraction is important here. If this function is proper, then we can assume that these partial fractions are likewise proper fractions. So the two proppers add together to give you a proper fraction. Well, if the denominator is a linear, x minus 3, and it is a proper fraction, the numerator needs to be a smaller degree. But the only degree smaller than x minus 3 would be 0, which is a constant. So it turns out the numerator is just a number. A number we don't yet know. We'll call it capital A. Same thing for this one. What's the numerator going to be? Well, if this is a proper fraction, then that means the denominator needs to have a bigger degree than the numerator, which only leaves the possibility of degree 0, which would be another constant. So the template of this partial fraction decomposition is going to look like A over x minus 3 plus B over x minus 2. It doesn't matter what you call the numerators. They're just going to be numbers. Oftentimes, we use capital letters to denote them here. All right. And so then we have this equation. This template gives us an equation. x over x minus 3 times x minus 2 is equal to A over x minus 3 plus B over x minus 2. The next step here to solve this partial fraction decomposition is we're going to clear the denominators. That is, we're going to take the equation above and we're going to multiply the left and the right-hand side by the least common denominator, which is x minus 3 and x minus 2. We do that on the left. We do that on the right. Now the left is going to be pretty easy. It just cancels out. And so we end up with the left-hand side just being x. This will be the numerator of the original rational function. On the right-hand side, you do have to distribute these things. And when you have A times x minus 3 times x minus 2 over x minus 3, what you're going to see here is that the x minus 3s are going to cancel leaving you just an x minus 2. So you get A times x minus 2. And then for the second one, you get B times x minus 3 and x minus 2 all over x minus 2. So you're going to see the same thing here that the x minus 2s are going to cancel leaving you just with the x minus 3. And so now the denominators are clear. Step 5 here. We are going to destroy values of x. We want to annihilate things. So destroy, right? What does this mean? This destruction phase, this annihilation phase means that you're going to pick values of x that make the denominators go to 0. What are you talking about? It's like dividing by 0 is the same thing as dropping an atomic bomb on SUU. Why would we do that? Well, notice that when we cleared the denominators, there are no more denominators. So we can actually plug in things that would previously made the denominators go to 0. So what that means for us here is that we're going to plug in, for example, x equals 2. That made this denominator up here go to 0. When we plug in x equals 2 into the equation, see what happens. You're going to get 2 is equal to A times 2 minus 2 plus B times 2 minus 3. And that simplifies to be 2 equals, well, you're going to get 0A minus B, which gives us 2 equals minus B. That tells me that B should equal negative 2. All right. So we annihilated A by plugging in x equals 2. On the other hand, what if we were to plug in x equals 3? x equals 3 would give us that 3 equals A times 3 minus 2 plus B times 3 minus 3. You can see what happens this time that plugging in x equals 3. The 3 minus 3 will go to 0. That disappears and you're left with 3 equals 1A, which is just A right here. So we found the coefficients for our partial fraction decomposition. And so then in the end, we get x over x minus 3 and x minus 2. This equals the A value, which was, remember, let's double check what we had before, right? A was above the x minus 3, B was above the x minus 2. So we end up with 3 over x minus, oh, I already forgot what it was, 3 minus 2 over x minus 2. And I better double check again to make sure I didn't break it. I hope that was the right one. Good. And so this then gives us our partial fraction decomposition. We were able to add together, that is, we found the partial fractions that add together to give us this original fraction right here. And you can check. I'll leave it up to the viewer here to check that 3 over x minus 3 minus 2 over x minus 2. That if you add these things together, you get back the original fraction. Let's look at another example here. Let's go through a little bit quicker this time. So we have 3x over x plus 2 times x minus 1. Is this thing proper? Right? Could this sit with the Queen of England? The answer is yes. Great. It's a proper fraction. The top is linear. The denominator is quadratic. Then the next step is to factor the denominator. We're done. With that step, that's great. So we move immediately to the template here. We have 3x over x plus 2 and x minus 1. So our template is going to look like a over x plus 2 plus b over x minus 1. We then cleared the denominators. We're going to get 3x is equal to a times x minus 1 plus b times x plus 2. You'll notice that the number that gets attached to the to the a right here is the number with the denominator was missing. And the same thing with the b, the number, the factor getting attached to the b was the one that was missing. So we get 3x equals a times x minus 1 plus b times x plus 2. Let's pick these annihilating values here. Let's take, for example, x equals 1. If you pick x equals 1, the left hand side becomes a 3. A would be annihilated and you'd be left with 3b. 1 plus 2 is 3 there. So b equals 1. That was pretty quick there. If we take x equals negative 2, that's another annihilating value. That would annihilate b in this consideration here. If you plug in x equals negative 2, you get negative 6 on the left hand side and you're going to get negative 3a, which means a equals 2. So our partial fraction decomposition then would be you get 2 over x plus 2. And then you're going to get 1 over x minus 1. So this gives us our partial fraction decomposition. And it's a really neat trick that as long as you have your template, you clear the denominators, you plug in these annihilators, and this will give you the numerators of this partial fraction. And you can check 2 over x plus 2 plus 1 over x minus 1 does in fact give us the correct partial fraction decomposition. Just add the fractions together and you'll see that you ended up with what we started with.