 In this video, we provide the solution to question number 14 for the practice version of exam four for math 1050, in which case we're given a rational function right here. 11x squared minus 8x minus seven is the numerator, and the denominator is factored as 2x squared minus one times x minus three. And so given this rational function, we want to compute the partial fraction decomposition, for which it's very fortunate for us that the denominator is factored, it saves us a little bit of effort. So note here that when we're trying to compute the partial fraction decomposition, the first thing we want to do is find the template. And the template is based entirely upon the denominator. I guess I should say the first thing to worry about is is this even a proper fraction? The numerator has degree two, x squared, the denominator has degree three. So this is a proper fraction, no divisions necessary here. So then we proceed forward to find the template for the PFD partial fraction decomposition, in which case looking at the denominator, we're gonna have something like the following. Well, the 2x squared minus one, it's a quadratic, so the denominator can be 2x squared minus one, and the numerator would be then some ax plus b. Given that you could factor 2x squared minus one using the difference of squares, it would be irrational, but you can do that if you wanted to have linear factors. I'm gonna leave it as this irreducible quadratic instead. And then the last one, the linear factors are our favorite, x minus three, then it's numerator would then be some constant c. So we have to then find these numbers. So the next thing to do is to clear the denominators. You're gonna times both sides of the equation by the denominator, 2x squared minus one, x minus three. You do that over there times my velocity on the left so they cancel out perfectly. Here, of course, you need to distribute. Some things will cancel, some things won't. We'll be left with 11x squared minus 8x minus seven. This is gonna equal ax plus b times x minus three. The thing that didn't cancel out was the thing you didn't have in the denominator. So the x minus three is still around. And then you're gonna get c times 2x squared minus one, like so. So we can continue to multiply things out and we can try to solve a system of linear equations. I'm gonna prefer to actually just use some cool values. That is, I'm gonna annihilate some things. For example, I wanna get rid of this x minus three. So I'm gonna plug in three. So we need to plug in x equals three on to the left-hand side and you can do that. Three squared is nine times out of 11, et cetera, et cetera. I'm just actually gonna use synthetic division because it gives you the evaluation and the arithmetic is generally much simpler since it doesn't actually require the calculation of exponents of any kind. So bring down the 11. 11 times three is going to be 33 minus eight is 25 times three is 75 minus seven is 68, like so. So we see that the left-hand side is gonna turn out to be 68. On the right-hand side, when you plug in x equals three here, you get zero, the whole thing is annihilated. So we just have to plug in three here. This one's not as complicated. So I'm just gonna plug and chug in the usual manner. So we get c times two times three squared minus one. Three squared is nine times two is 18 minus one is 17. So you get 17 times c. We wanna divide both sides by 17, like we did right here. And then 17 goes into 68 exactly four times. So that was fortuitous. We get c is equal to four, great. The next thing we wanna do is because we have this thing factored right now, I can't really plug in something that's going to just make it disappear unless I wanna stick in an irrational number, which I probably don't wanna do. So what I'm gonna do instead is play off of the fact I have a zero right. I could plug in x equals zero for the a and that would then kill off the, I could plug in zero for x right here, kill off the a. Since I know c, right, I mean I can come back here retroactively fixing this thing right here. We have a plus four like so. In fact, if I wanted to, I could even move all this stuff over here, combine some like terms that sounds kind of fun. You're gonna take 11x squared minus 8x squared that's gonna be a 3x squared. Then we have a negative 8x that didn't change. You're gonna have a negative seven. We have a minus four, so we add to this, plus four we're gonna have a minus three right now. And so this is equal to ax plus b times x minus three like so. So again, I could plug in the x equals zero and go for here. It does turn out though on the left-hand side though, if you wanted to, you actually could divide by x minus three. So I'm just showing things a little bit different than how we usually do it. You actually could divide both sides by x minus three, right? Just providing an alternative method here. So again, pulling out synthetic division, three negative eight, negative three. If we divide by three here, what happens? Bring down the three, three times three is nine. Minus eight is one times three is three. I gave you a zero just like I said it would be. In which case then the left-hand side actually simplifies to be three x plus one equals ax plus b. So without any plugging whatsoever, we see that a is gonna equal three and b has to equal one. So again, alternative way of approaching this one. So then coming back to the final four and I'll just come back up over here since I kind of ran out of space. Our final answer is then gonna be the following. A remember turned out to be three. So we get three x, b was a one over two x squared minus one. And then for the second part, c turned out to be a four that's still on the screen. Of course, we get four over x minus three. And this then gives us the partial fraction decomposition of this rational function.