 Let's factor completely the degree 5 polynomial f of x equals 2x to the fifth plus 5x to the fourth minus 8x cubed minus 14x squared plus 6x plus 9. If you look at variations of signs, I see one variation here and two variations, the second one right there. So the variation of signs tells me that there's going to be either two or zero positive roots. If you switch the signs for the odd degree powers, so you get negative, positive, negative right there, you're going to see one variation, two variations, three variations, and you see that the variations often add up to the degree there. And so the negative variations is going to be three or one. And so with this in mind, we have two positive roots or none. So there could be none if we look for them. When it turns out, when we look at the negative roots, there's going to be three or one negative root. And so the thing is we do have a guaranteed negative root. But the one thing you have to remember about Descartes' rules, Descartes' rule applies to reels. It doesn't actually guarantee that there's going to be a, there's doesn't going to be guaranteed there's going to be a rational negative root. It could be a real root. And so we have to be careful and we can only use this so much, right? This does make me kind of think that I'm more likely to find a negative root if I go looking for those first. It's again, it's not a requirement, but you know, you might want to consider it. Looking for negatives first. But in this situation, I don't see much benefit of doing one over the other. Because honestly, there could be two real, two positive roots, only one negative root. So there might be more positives. So this one's kind of a mixed bag right there. Descartes' rule sometimes is. I really like Descartes' rule when it tells me that there's no positive roots or no negative roots. That can be very helpful. That's when it's the most helpful. So if we look at the, using the rational roots there, we have to look at the factors of nine and divide them by the factors of two. And so the possible rational roots we see here, so we're going to get plus or minus one, plus or minus three, plus or minus nine, the factors of nine. We have to divide those by one, which we already did. We have to divide those also by two. So we get plus or minus one half, plus or minus three halves, and plus or minus nine halves. Now there's one thing I want to tell you here. If you're leading coefficient is anything other than one. One of two things have to happen. Either every coefficient is divisible here by two, which it's not. Or we have a root, which is not a whole number. So that means we're going to have to look beyond one, three, and nine to find the roots of this thing. Now it very likely could be that one of these fractions turns out to be a root or it could actually be that we have some non real or irrational roots. Those are possibilities as well. So if you a poor fractions, the point is we're going to have to probably try them eventually. But I can understand that we might hesitate to do it at first. So let's try, for example, let's try looking for, let's try one. I like to try one. One sort of a good start, I think, and we can see where it goes from there. So after all, one kind of is in the middle, because three and nine are bigger than one. So is nine halves and three halves. We won't have a smaller. And again, I'm trying to avoid the fractions here. Let's just try one. See what happens. Two, five, negative eight, negative 14, six, and nine. We are going to divide it by one is what I said we're going to do. So bring down the two. One times two is two plus five is seven times one is seven minus eight is negative one times one is a negative one minus 14 is a negative 15 times one is negative 15 plus six is going to be giving me a negative nine times one is negative nine plus nine is zero. Oh, great. We found we found a root on our very first try. That's wonderful. So I guess it's a factorization f of x is going to factor as x minus one times two x to the fourth plus seven x cubed minus x squared minus 15 x minus nine. And so you look at things here of what changed, right? Our constant term is still and well, it's which from positive to negative nine doesn't make much of a difference. The leading coefficient is still a two. So the list of potential rational roots doesn't change whatsoever. In terms of variation of signs, let's see here, we have now one variation of science that means there's one positive root left. That's not surprising because the options were either two or zero. We found one. So that has to be a second positive root. Okay. It could actually be one, right? The fact that one worked once doesn't mean it can't work again. It could be a repeated root. So it is worth our time to check what happens if we try one, but now we have to use the quotient. Don't use the original one anymore. Once you find a remainder, always switch to the quotient. Two, seven, negative one, negative 15 and negative nine. So let's try one again there. And so doing that, we bring down the two. Two times one is two plus seven is nine. Times one is nine. Minus one is eight. Times one is eight. Minus 15 is a negative seven. Times one is negative seven. Minus nine is a negative 16. And so we do see there's a bunch of positive and negatives right there. So in terms of our rational roots, we've now exhausted one as a possibility. One is no longer going to work. So let's kind of record what we have here. So what's remaining is we could still have a negative one. That's a possibility. We have plus or minus three. We have plus or minus nine. We also have some fractions, plus or minus one half, plus or minus three halves, and plus or minus nine halves. These are possibilities still. So since one didn't work, and you see it's off the list now, I'm going to erase these numbers and try this again. Once I have it recorded that one no longer is going to be, there's not a repeated root. I don't need any information from the synthetic division. Let's try say three now, the next one, because we need to find the next positive number. Let me try three. Bring down the two. Two times three is six, plus seven is 13, times three is going to be 39, minus one is going to be 38. 38 times three is 114. We can see where this is going. Minus 15, you get 99. Three times 99 is going to be 297, minus nine is going to be 288. You can see that all the numbers on the bottom are positive. This is an example of the upper bounds theorem. This tells us that three is too big. Anything bigger than three needs to be taken off the list. Plus three doesn't work, but that also means that plus nine doesn't work either. What about nine halves? Is that bigger than three? Nine halves would be four and a half, so yeah, nine halves doesn't work either. In terms of positive roots, we still have one half as a possibility and three halves as a possibility. As much as we might not like it, it's worth checking one of these fractions. Which one is it? When I did one, I didn't get that one was an upper bound, so it could be that something is bigger than one. If I have to choose between one half and three halves, my inclination is, well, it needs to be less than three, but maybe bigger than one, probably bigger than one. Not a guarantee yet, though. I'm going to try this using three halves. What does it look like if we try three halves? Again, I'm going to erase this since we now have an upper bound. Let's see my potential rational roots right here. Let's try this again. Third try. We're going to try three halves. That's what my gut tells me here. Bring down the two. Now, this is just going to be a multiplication, so multiplication by fractions. Two times three halves is going to give me a three. Three plus seven is equal to ten. That's nice. Ten times three halves, notice that two goes into ten five times, so you get three times five, which is a 15. A 15 takeaway one is a 14. That's another even number. This is nice. Two goes into 14 seven times. Three times seven is 21. 21 takeaway 15 is six. And two goes into six. Three times three times three is nine. When Bob's your uncle there, negative nine plus nine is zero. So we found our root. That's wonderful. And so now let's look at the factorization. F of x then factors as x minus one. We then have x minus three halves as a root as a factor here. And then the other one looking at the quotient here, you're going to get two x cubed plus 10 x squared plus 14 x plus six. And so this gives us a factorization. And as much as we didn't like it, maybe we have to we have to accept that three halves was in fact a root you get fractions sometimes. Now look what happened here. You know, you have this x minus three halves. If you don't like fractions, get you're going to like the next step here. Notice that if we look at the quotient here, two x cubed, 10 x squared, 14 x plus six. There's two things I want to point out here. There's no variation of signs right here. The variation is zero. That means there's no more positive roots. But that matches up with what we saw originally. The original variation was two. We've now found our two positive roots. There's one and three halves. So there's no more positive roots. That affects things a lot. Also, notice that each and every one of these coefficients is positive. Two, 10, 14 and six. We could factor out a common factor of two. If we did that, we would get two times x minus one times x minus three halves times x cubed plus five x squared plus seven x plus three. That's a that's a critical observation here because that affects our rational roots test here. If we were to run the rational roots test on this guy right here, notice the potential rational roots were going to be divisors of three, which there's no positive divisors left. So we want to be trying negative one and negative three. That's a lot better than six here because there's going to be some things that we don't want to try. So we want to factor out as possible. Also, if you don't like fractions, notice you could distribute the two onto the x and to the negative three halves here. And that would give us the factorization x minus one times two x minus three times x cubed plus five x squared plus seven x plus three. And so that's pretty nice fact, nice factorization. And the factorization has no fractions in it whatsoever. What we see here is an example of something called Gauss's lemma that when you factor a polynomial, which if it originally had integer coefficients, then the factorization can be done with integers as well. You don't need fractions because even though we got we got three halves of root, you're able to pull off a factor of two to kind of clear out the denominators right here. It's like providence that provided for us. This is actually result proven by Gauss here. So now we want to look for factors of our cubic quotient here. So we want to either negative one or negative three. Whichever one you prefer, don't matter. It's sort of a coin toss at the moment. So let's just try negative one. So take your coefficients one, five, seven, and three. Clearly that was way too long of a line. Let's try negative one here. Bring down the one. One times negative one is negative one. Plus five is four times negative one is negative four. Plus seven is three times negative one is negative three. Which gives us, turns out negative one was a good guess right there. And so now we're ready to find our factorization here. So we're gonna have f of x equals x minus one times two x minus three times x plus one. That was the factor we just did here. And then what's left over is you get x squared plus four x plus three. Which that's a quadratic. And I noticed that factors of three that have to be four would be one and three. So I can factor this one more time. So just factor the quadratic, keep down the things I have. Once you have it to a quadratic, factor it how we have been doing it. In which case you're gonna get x plus three and x plus one. So it turns out it didn't matter which one you tried. Negative one with negative three, they would have both worked. But also knows that x plus one shows up twice. So I'm gonna actually write this as a repeated root. So you get x plus one squared. And so this gives me the complete factorization of f of x. It's x minus one times two x minus three times x plus one squared times x plus three. And so if we set this equal to zero, we can then solve this equation and get x equals one, three halves, negative one and negative three. Just like the rational roots theorem told us it would be.