 In this video we provide the solution to question number 25 for the practice final exam for math 1050. We're given a polynomial equation, well function I should say 2x to the fourth plus x cube minus 35x squared minus 113x plus 65. We have to find all complex or real zeros, all the real roots or complex roots for this one here. All right, so we want to start factoring this thing. Let's see, what can we do? So notice by Descartes' rule of variation of signs, there's two variations there, so that tells us there's either two positive roots or no positive roots, right? That would also assume a calculation would tell us that there's two negative roots or none, right? In which case there could be some complex roots that are not real, so we should be cautious about that. That doesn't really tell us too much. It's about equally likely there's positive or negative roots. So if we apply the rational roots theorem, what are my possible rational roots here? So looking at divisors of 65 divided by divisors of 2, 65 of course factors as 5 times 13. And so your possible divisors of 65 are plus or minus 1, plus or minus 5, plus or minus 13, plus or minus 65. We have to also divide these by 2. So we have plus or minus 1 half, plus or minus 5 halves, plus or minus 13 halves, and plus or minus 65 halves, like so. And so we then have to find which of these roots is going to work. You never want to try the big ones because if a big one worked, it meant a little one works as well. So try smaller ones first. There's a little bit of guessing and checking. So notice here that we have some fractions as possibilities. And in some regard, we might have to go there eventually, right? Because when you look at this polynomial, the leading coefficients 2, but none of the other coefficients are divisible, too. So the only way that 2 could be the leading coefficient is that you have a non-integer root. It could be a rational one, such as these ones right here, or it could be irrational or non-real as well. So at some point, we're going to have to do something that's not an integer. But we don't necessarily have to start there, right? We could try a couple of things. Again, I kind of want to hesitate to do the fractions until I have to. So I might try one or five, you know, something like that. You don't want to try something too, too big. So we could try one because there are some numbers less than one, like one half is less than one. But all these other ones are bigger. So I'm actually, I want to try something kind of in the middle so I can try this hot and cold game. So let's try positive five. We're going to do synthetic division. So make sure you have all of the coefficients there. 2, 1, negative 35, negative 113 and 65. And we said we're trying five. Go through the calculation there. Bring down the two. Two times five is 10 plus one is 11. 11 times five is 55 minus 35. That's going to give us 20 times that by five. We're going to end up with 100 minus 113 gives us negative 13. And then five times 13, like we said, was 65. So actually that was a pretty good guess to start off with. We ended up with 65 minus 65, which gives us zero. So now we have a factorization. We should always use the factorization when we find it. So we're going to get f of x equals x minus five times 2x cubed plus 11x squared plus 20x minus 13. So let's reevaluate what we have right here. We found one of our positive roots. So if you look at the variation of signs, you only have one variation. So there's going to be one positive root left. There's either two or zero negative roots in terms of the possible rational roots. Take p over q for a moment. So factors of 13, you're going to get plus or minus one plus minus 13. We haven't tried those ones yet, but you also have these fractions here. One half, one half, 13 halves. And then that's needed on this one, right? So we eliminated a lot of the fractions as a possibility. Five halves and 65 halves no longer work. Five's now out from consideration. So which of these remaining ones do we want to try? Again, at some point, we probably should try a fraction I don't want to yet. So let's try x equals one. That seems like a good choice there. So if we take the coefficients two, 11, 20, and negative 13, if we try one in that situation, bring down the two. Two times one is two plus 11 is 13 times one is 13, plus 20 is 33 times one is 33 minus 13 is 20. That was not a root, but notice that everything was positive on the bottom. Turns out one is an upper bound. So let's eliminate some things. One did not work. Also, 13 will not work because it's bigger than one. 13 halves will also not work because it's bigger than one. And so the only positive rational left is one half, which I know there's one more positive root. It could be irrational, so it might not be one half. But like I said, that coefficient of two right there suggests we're going to have to have something that's not an integer. So it seems like one half is worth a try. We've been avoiding it up until this moment, but it kind of feels like it's the last positive root that we it's worth trying here. Because if one half doesn't work, it means the other positive root is irrational. Let's try it. So two times one half is one plus 11 is 12 times one half is six plus 20 is 26 times one half is 13, then minus 13 plus 13 is zero. So we did find it after all. So my hesitancy, although it was valid in the end, it turned out it was going to work. So let's see what we have now. f of x is equal to x minus five, that's what we have. Then we have x minus one half, then we're going to get two x squared plus 12 x plus 26. Notice that everything in the quadratic is divisible by two. We can factor out that two and actually then redistribute it onto the x minus one half there. So we get a much better factorization x minus five, we're going to get two x minus one. And then our quadratic is going to look like x squared plus six x plus 13 like so. Looking at that polynomial x squared plus six x plus 13, there's no variation of science. There's no positive roots. The remaining roots are either going to be both negative. But again, they could be irrational. They could be or they could be non real. But the good news is we have a quadratic polynomial. We could find the remaining roots by just simpler factorization processes, right? Factors of 13 that have to be six, that's not going to happen. So there are actually are no more rational roots. We should try the quadratic formula. x equals here, we get negative six plus or minus the square root of 36 minus four times 13 is going to give us 52. And this all sits above two here, 36 minus 52. We see that discriminant is going to be negative. So we actually have some non real solutions here. We end up with a negative 16 over two. The square root of negative 16 is going to be negative four I factor number two from the top, you get negative three plus or minus two I there all over two, the two is cancel. And so we see that the remaining roots are going to be negative three plus or minus two I. And so we've now found all four roots of our polynomial. We have x equals one half, five, and then we have negative three plus or minus two I as our four roots. Two of them turned out to be non real. That's okay. The instructions wanted all complex roots. And there they are.