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Use Descartes' Rule of Signs to determine how many positive and norr many negative real zeros the polynomial can have. Tnen determine the possible total number...

Question

Use Descartes' Rule of Signs to determine how many positive and norr many negative real zeros the polynomial can have. Tnen determine the possible total number of real zeros (Enter your answers comma separated lists.) P(x) = x5 3xnumber of positive zeros possiblenumber of negative zeros possiblenumber of real Zeros possible

Use Descartes' Rule of Signs to determine how many positive and norr many negative real zeros the polynomial can have. Tnen determine the possible total number of real zeros (Enter your answers comma separated lists.) P(x) = x5 3x number of positive zeros possible num ber of negative zeros possible number of real Zeros possible



Answers

Use Descartes’ Rule of Signs to determine how many positive and how many negative real zeros the polynomial can have. Then determine the possible total number of real zeros.
$$
P(x)=x^{5}+4 x^{3}-x^{2}+6 x
$$

This question is essentially asking us to use dick carts rule signs to determine how many positive and negative rials zeros the polynomial can have. What this means is that we're gonna be setting acts, is equal to negative acts, solve for P and negative acts. We get two extra. The sex was five x the fourth plus X cubed post five acts minus one. Okay, now that we have this, let's count the sign changes between the terms of P o box. We got one sign change. Therefore, we have one positive real zero now countersign changes between the terms of p 90 x. We get one sign change, which means one negative reels here. Therefore, we conclude that there are two riel zeros.

In this video we're going to use to cultural signs to see how Maney positive rials heroes there are How many negative rials years there are. And how many total rials here is There are so to start out, we're going to make a little chart down here in the bottom. Okay, Where can do a plus for the positive here, minus for negative. And I'm right. T o t my abbreviation for total. Now notice. Here we have a This is a degree six polynomial, which means that the total number of rials zeros we get is going to be less than or equal to six. Okay, we're not gonna have more than six. Okay, now let's do this. So theorem states that we need to look at the change in signs. Okay, so I'm right a plus here, because that's a positive too. Okay, so, um, we're gonna start off by looking at the seeing. How Maney Positive riel. Sierra star. So we go from a positive too. £2 of five. OK, so that's no change. Okay? And then we get a 52 and plus two negative. So that's gonna be one change. And we got from negative to negative that zero. And we go from negative to negative. That is also zero. So there's one possible rial. Positive. Great. Okay, Now, to do the negative one we're going, Teoh, plug in instead of an X, We're gonna do negative X. So it's gonna be too. Times negative X to the sixth, plus five times negative X to the fourth minus a negative. X cubed minus five times negative X minus one. Okay. And so now let's see the change. Okay, so we start with a negative X to the sixth eyes. Just gonna be equal to X to the sixth. So right now we start off with a posit that's going to be positive and then negative x to the fourth. It's negative. X times negative. X times negative X times negative x just could be a positive X. So this is a positive plus 5/4 uh, negative X to the cube. Negative. X to the third is equal to negative x of the third. Uh, but we are so tracking in negative between we're adding Okay, so that's a positive. Minus five times negative, X. It is also a positive cause. We're subtracting and negative. And then negative one. That's a minus. So, looking through here, we have no change. We have no change. New change. And there we go. We have one change here. Okay? Which means the total number of riel zeros for this polynomial here is to Okay, now, just a quick side note on the explanation here. So too is less than or equal to six. So that's good. But one thing to note is that whatever decree the polynomial is, so this is a degree six polynomial. That means it's gonna have six routes, but Onley two of them are really meaning four of them are complex or your or imaginary. But we're going to get to that and later videos. But for now, box and red down here. This table, this chart, it's going to be your answer

This problem. Ask us to use the cards signed rule to determine the number of positive real zeros and the number of mhm real negative zeros. So let's start by positive. So this would be plus five. So nothing here there is one. There's two, and that's also positive. So we're going to have to. Or this could be to minus a an even number minus two, so it could be two minus two, which is equal to zero so that positive real zeros is either two or zero. Now let's check the negatives. So we just put F of negative X in here and then see what these signs turn into. So let's be negative. Exit Power 55 Negatives. This would stay as plus three extra power. Four. This would be negative times negative. We tried to make it plus X cubed. This will be minus two X plus three. So let's count them up. So I'd have one here, one here for a total of two. Which means that the negatives could also be two. Or am I missing one? Sorry, I'm missing one right here. So the negatives. That's three. The negatives could be three or three minus 23 minus two. Which equals one, right? You just subtract by that they even the next even. Aren't some even integers positive? So one for three or one for the negative two or zero for the positive.

This problem as you used the card sign rule to determine how many possible Rio zeros there are for positive and negative real zeros. So we're looking for flip flops and signs here, so positive to negative everyone, there would be two. There will be three. There'd be four, so it could be four. Now, it could be four or four minus four and four minus two and no particular order. You gotta subtract or even number. So before, or zero or two. So four or zero or two real positive zeros. Now let's look at the real negative zeros. So this becomes valuable for F of negative X. This is negative X to the power five, which would leave this whole thing as negative two x to the power five. This would stay as negative extra power for because this would be positive extra power for one negative in front. This becomes minus execute because this becomes negative. Execute. And then this becomes minus X squared because this stays as X squared with one subtraction in front minus and X plus five. So let's see if there's any flip flop in here. Nothing. Nothing. Nothing. Nothing. Nothing. One at the end. So this has a real negative zero of one


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