 In this video, we provide the solution to question number 16 for practice exam number three for math 1050. We're given the degree four polynomial f of x equals x to the fourth minus 5x cubed minus 5x squared plus 23x plus 10. And we need to find all of the real or complex roots or zeros of this polynomial. So the first thing I always recommend when you're working with these is to consider the rational roots theorem and what are the possible rational roots. So we get that p over q. They could equal looking at the constant term and dividing it by the leading coefficient. The leading coefficient is just a 1 here. So we only have to look at divisors of 10. So we're going to have plus or minus 1, plus or minus 2, plus or minus 5, and plus or minus 10. When it comes to these things, never try the big ones first because if a big one like 10 worked, it also would mean that 1 would have to work as well. And so therefore, big ones only work when small ones also work. But sometimes big ones don't work, but the smalls do. So try small values there. We could try Descartes' rule of signs, variation of signs. We see one variation right there. We see two variations. So there's two possible positive roots. That's a possibility. And as there's four roots total, that makes me think that there's probably two negative roots. I mean, I should check it specifically, but I'm not seeing an obvious thing. Really what Descartes' rule for me is nice is when you have no alteration. That means there's no positive roots. I love that. When you alternate each and every time, positive, negative, positive, negative, positive, that means there's no negative roots. I love those. And so unless I see that, I usually don't need to use Descartes' rule too often. So I need to try something. I'm just gonna start off with positive one, see what happens there. So we're gonna do synthetic division because we have to check to see if one is a root. And the nice thing about synthetic division is actually instead of function evaluation, it actually is faster at checking if it's a root. And if it is a root, we find the quotient at the exact same time. So it really is the most efficient way to do it, use synthetic division. Bring down the one. One times one is one, minus five is negative four, times one is negative four, minus five is negative nine, times one is negative nine, plus 23 is 14, times one is 14, plus 10 is 24. So we see that this is not equal to zero. So one is in fact not a root, right? We looking here at the denominator, one was a positive number. It could be an upper bound for all of the positive roots. But of course we don't see all positives down there. So that means it's not necessarily an upper bound. So that makes me feel like there's probably a bigger number we could try. But after all, one was the smallest possible rational root we could have had there. Of course there could be irrational roots we find later. So that's not to say that there couldn't be another one, less than one, but we're trying these rational roots first. So I'm inclined to go try two, because again, DeCartwell says there could be two positive roots or there might be none. We could be searching in vain here, but we have to do a little bit of experimentation. So let's try again using two. This is one of the advantages of synthetic division is that even though we have to experiment, since synthetic division is very efficient, it's not too costly to do the experimentation here. For question number 16, we have the whole back page of the test to do it. So feel free to do several of these divisions as you try to find the roots. So bring down the one, one times two is two, minus five is negative three, times two is negative six, minus five is negative 11, times two is negative 22, plus 23 is one, times two is two, we end up with a 12, right? So that one didn't work either. And so again, looking at the denominator, I don't see all positives. This still has positive and negatives, which means there's a positive root and it's probably bigger. So not knowing what the irrational options are, I'm gonna then try five, right? So again, negative one, negative five, negative five, 23 and 10, try a five this time. Bring down the one, one times five is five, minus five is zero, times five is zero, minus five is negative five, times five is negative 25, plus 23 is negative two, five times negative two is negative 10, and Bob's your uncle we found a root that time. That's excellent, great, great, great, great. Whenever you find a root, you have to of course do a short celebration, you go woo, woo, you're excited because we found it. And then we want to then use that to factor the polynomial. So what do we know so far? So f of x right here, now that five is a root, we have x minus five, that's what we discovered. Then looking at the depressed polynomial right here, you're gonna get lower the power by one, so we go from degree four, now it's gonna go down to degree three, so we get x cubed, plus zero x squared minus five x minus two, like so, and then what we're gonna do is we're gonna basically start the whole problem over again, I mean we still use information we know, but like let's do the rational roots test again, right? So we have p over q, what are the possibilities? So the leading coefficient still is one, that didn't change, but now the constant term is negative two. So we have to look at factors of two, factors of two that, divided by factors of one. So we're gonna get negative one and negative two. Now notice what I did here, normally when we do the rational roots test here, we're gonna have plus or minus one, plus or minus two, but we've already tried positive one, and we've already tried negative two, or positive two and they already failed. If it failed for the original polynomial, it'll also fail for the depressed ones because the factors of the depressed polynomial is also a factor of the original one. So positive one and positive two don't work, we don't know about negative one and negative two yet. Now, if we look at Descartes rule of sign, there was one variation of signs. So that does tell us there was a positive root, and from our previous list, we had positive one, which failed, positive two, which failed, positive five, which passed, and then 10s no longer in consideration. So even though there is one more positive root, it's gonna probably be irrational, which is why it's not on our list, because again, five and 10 are no longer contenders and one and two are already ruled out. So let's look for the negative roots that are possible here. Not sure which number to use, negative one or negative two, it's kind of like a flip of the coin. Let's just try negative one and see if we get lucky here. So you have one, zero, negative five and negative two, you write down the coefficients, and then we're trying negative one here. Bring down the one, one times negative one is negative one, plus zero is negative one, times negative one is positive one, minus five is negative four, times negative one is positive four, minus two is then a positive two. Two or negative one didn't work, we didn't get a root. So remove that from consideration. Well, with no other options, it looks like negative two, cross your fingers, this better work, otherwise we're in our boat without a paddle. Negative two is gonna work, trust me on this one. Bring down the one, one times negative two is negative two, plus zero is negative two, times negative two is four, minus five is negative one, times negative two is positive two, minus two is zero. So it worked in that situation, whoopee. So then factor your polynomial one more time using the information we have. So we have X minus five was one of the roots. We have X plus two is the other factor, because negative two just worked. You always wanna write the factorization. We're looking for roots, but we could have repeated roots. Those are things we have to watch out for. So don't just list the roots, list the factorization, you're gonna be much better off that way. And now looking at the depressed polynomial, we have X squared minus two X minus one, like so. Now the depressed polynomial has reduced to a quadratic polynomial. So we can just factor using the quadratic factorization techniques we know about. So we need factors of negative one that add up to be negative two. The only way to factor negative one using integers is negative one and one which adds up to be zero. So it turns out that there's no rational factors left, which is not surprising. The only one who possibly could have worked was negative two, which by the rational roots test isn't gonna work here anyways. So the remaining solutions have got to either be non real or irrational. So we should use the quadratic formula in this situation. So X equals plus or X equals negative B, so that's a positive two, plus or minus the square root of, now we have B squared, which is four, minus four AC. So you're gonna get negative four times one times negative one, which is actually positive four right there, all over two A, which is two times one, which is two. Simplifying this a little bit, we got two plus the square root of eight over two. Eight of course is the same thing as four times two. Four is a perfect square, so we get two plus or minus two, whoops, two root two over two. Since the numerator, everything's divisible by two, you can factor out the two and cancel in the bottom. We're now in the situation that we're looking for. Our roots were five, negative two, and then we have these two irrational roots, one plus, one plus or minus the square root of two, like so when we continue to reduce this thing, again factor out the two there. You get the four roots, you can write it as one plus or minus root two, or you can write one plus the square root of two, one minus square root of two. I don't care either way, but these are the four roots of our polynomial. So this last exercise is definitely the most lengthy, and the length of it will kind of depend a lot on the trial and error you go through, but using these strategies like the synthetic division, upper and lower bound theorems, rational roots theorems, Descartes rule of variation of signs, these can help speed up the process, so it's not just random guessing. You can be strategic as you do this phone book search to find the roots.