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Question positive real zeros of P(x) =5x2_2x+3 Use Decartes'Rule signs to determine the number of possible0A B. 2 c5 D.4 OE 3...

Question

Question positive real zeros of P(x) =5x2_2x+3 Use Decartes'Rule signs to determine the number of possible0A B. 2 c5 D.4 OE 3

Question positive real zeros of P(x) =5x2_2x+3 Use Decartes'Rule signs to determine the number of possible 0A B. 2 c5 D.4 OE 3



Answers

Determine the zeros of the polynomial $f(x)=x^{3}+(b-a) x^{2}-a b x$ for the positive real numbers $a$ and $b$.

Hello for this problem we are going to be using discarded rule of science to determine the total possible number of real zeros for our polynomial function now discriminates rule of Science says that we can use the change of science between the terms and our polynomial function to help us figure out how many positive real zeros are and how many negative real zeros there are. So going ahead and looking at this function that we have. Um Let's look and see how many sign changes we have. So we have a sign change from here to here. Sign change from here to here and a sign change from here to here. So we have a total of three. Sign changes between our terms. Now discard the rule of Science says that we can have either one of two things here. We can either have the same number of positive real zeros as we have signed changes or we can have the number of sign changes minus a positive even integer. Um So It's a good thing for us is there's only one positive even integer between three and zero. That's too. So we take three minus two equals one. So essentially what we have is we can either have three or we can have one real positive zeros for this function. Yeah. Now to find our real negative zeros. What we do is we take our function and we plug in negative X everywhere we have an ex we take F of negative X. So I already have that written out here. Let's go ahead and simplify it down. That's going to give us a positive two X cubed since we have a negative times negative a positive X squared positive X and positive seven. So because all of our terms are positive, we don't have any sign changes. That means that we have no real negative zeros for F of X. Mhm. So putting all of that information together, We have either three or 1 positive real zeros and no real negative zeros. And there you have it.

Hello, everyone. So here we have Apolinar meal off degree four. So the first part we have to you don't mind the number off negative and positive real zeros by using the this Curtis true love science. OK, so in this method, what do you do is ah, we see how many vacation off science are there in the consecutive terms of the bull in a meal. So here you can see there is one variation. Okay. There is one variation there is also in violation. And Leslie, we have another variation. So the number of aviation gives the number off Positive real numbers, real zeros number off. Positive rial. Zito's would be the number of variations which are four. Okay, or we have. And even in digital, less than this variation. So what, even indigenous could we subtract from four to get other possible our number off real zeros so you can subtract two. So if we subtract, we will get to And then we can also subject four. So we'll get to zero. That means there are three possible number off native reels. It was either all the four would be negative real, And our two there will be two negative residents or none of them would be negative. And really? Okay, So these are the possible cases, not a C for the positive. Oh, says this is the positive real zeros. Number off was a difference of this. Okay, now let's check for the number of negative. Real serious. Okay, So number off negative, really? Zeros. So to find the number we have to consider Ah, be off Negative, X, get so p off. Negative X. In this case on leader terms containing our power will change their sign. So that would be positive. Positive then. Positive, positive and positive. That was There is no variation in sign. All are positive. That means number of negative real zeros would be zero. There will be no negative real zeros. Okay, fine. So now let us use the rational zero tear, um, to find the possible rational zeroes off this polynomial. So this is our next word, which we'll we'll be using that rational Macedo, your um so to say is that if we take the issues off the factors off all the all the factors off, the less term to the factors of the first room, where it here it is one positive one. So if we take the ratio of this two factor the factors off this two terms than we will get the possible zeros. Rational zeros. So let us, right. Ah, the factors for us. So the less term if we denoted is it all? And the first term coefficient is in, then is it is four and n is one. Okay, Some factors of four are in desert. Factors up wasn't negative. One was native to and positive Nearly four. And the factors off in a positive negative one only. So that issue a not by A M would give us the possible zeros. So possible congressional zeroes would be their issues, so their issues would be similar to the factor. So four. So that is okay. Now you can see as we have no negative really zeros. So since they have no negative real zeros, therefore we get right possible national zeros after applying the the Scotus School of Principal plus rational zero tear, Um, we can write possible rational zeroes would be Onley the positive values 12 and four. Okay, so let us check with one first. Somehow the part c which is testing. So let's check with one Soto, find whether one is zero. This pollen email. We will divide the pollen Immel synthetically by one. So to divide, we have to use the conditions. The conditions there. One negative too. Five minus eight and four. One negative, too. Five negative. Eight and four. Okay, so negative one will come down and we will multiply. Wanted negative one and put it here. Okay, so they will add up to negative three and one with negatives to be negative. Three and five ministries to okay, so one would multiply. That too is so is too negative. It positive too is so let me check. There is a problem, I think. One minus 25 minutes. It plus four. Okay, so these are Ah, fine. So if we divide by one, so negative one will come down. Yeah, but this is not negative on. This is positive in the first job. Okay, so this is the mystery here. One positive one will come down. So that means this will be positive. One. So negative two plus one is so we have negative one. Then one multiplied with negative one is negative. One and then five minus one will be four and four multiplied with one will be for negative. Eight plus four is negative food than one multiplied Negative four is before. So we got zero. So this part is the caution. And this part is the remainder. Okay, so that means one Is that zero of this polynomial second right? Zero one of the series one. So since we have one positive really zero, that means we can exclude this case. Okay, So we can have either two or four. Don't let us check again. So will see for two. So check for two. We have to use the remaining coffee ships. This one So one negative. 14 negative four So one will come down. That means ah do here. So negative one So a positive one than to multi brand to six. 12. So this is not a goes to zero, not it goes to zero this part So we will have Ah will not have to as the zero off this planet. So now let us check with the last one, which is four. So these are the national zeros on professional zeros. So for go straight for so one will come down for multiplied with one is for Okay, so then minus three, then minus trail and minus eight. And negative 32. So this part is negative 36 which is not equal to zero. That means four is not also the zero of this political. Okay, so then how come is this is possible that only one ah zero we have found which oneness positing. That is one because there are only 32 possible cases now that two of them would be positive or four of them would be positive. So since ah, one we've found. So there must be another If we consider this case so we can again check with one for the remaining part. Because we can have ah, duplicate zeros. Double zeros, isn't it? So you can check with one. So if we put one over here, So when will come down? Negative than positive. 10 said one multiplied 2004 then four. Yes. So we have found zero okay as a remainder for again one. So that means another zero off this polynomial would be one on Lee. So this one is multiplicity too. Okay, Multiplicity is to this this repeating zeros. Okay, now what we have were left to it is the coefficient of the remaining part. Okay, so here this condition will be written as the factor, which is X squared plus zero decks plus four. So that is equal strikes a squared plus four. So there's the remaining factor. Okay, so now let us jump into the last part. Okay, So these are the two is the only two rational zeros and really positive zeros we've found for this polymetal. Okay, so now let us some of the less. But where we pretty present p X s productive. It's ah, factors, linear factors. So since the factor term says that if is the zero off a pulling a meal, then x minus is a factor. Okay, so then here we have one as the zero and both to Buzios that there were general one. So that means X minus one. Then again, X minus one. So this stood with a factor. And the remaining factor was this one. So we can write it Excess square plus four. So this part cannot be reduced. Irreducible uh, civil. It is irreducible factor. Eso we cannot have real factors find. So that means we can write it. X minus one whole square. Multi bladed excess squared plus four. So this is the product form off this Polina meal in terms of its linear factors and irreducible quadratic sectors. Okay, so I hope you have understood this problem. So here the zeroes are only two zeros. That is both are equal. Onda, Multiplicity stooped and hear the case which is valid, this one. So to really positive zeros and no negative real zeros. So this is also invalid. Okay, so I hope you have understood. Thank you.

Hello, everyone. So here we have a politic male off degree full. Okay, so this one So the first dusk is true. Use the that's Curtis rule off sign to determine the number of positive real zeros and the number off negative real zeros. So I am writing here Will be changed the color first. Okay, so number off positive rial zeros and we have the number off Negative. Really? CDO's Okay, so let us find out. So using the discourteous rule off sign, we have to identify the variation in sign among the consecutive term off the given. Minimal. Okay, So for the positive real zero, we will use P X and for the negative real zeros we will use be off Negative X. Okay, So let us see. What are the vacation in science. So the first time, it's positive. So there is one variation inside positive to negative. Then again, we have another variation negative to positive and again positive to negative, then negative to positive. So there are in total four variations over here. So a number of positive real zeros are equal to the number of variations or ah, a number which is Ah, even indigent. Less than that number off really Zeros. Okay, so what is our other two? Even integers that we can subtract from this four. Okay, so one is two and other this for yourself. So if you subtract two so we'll get to and then if we subtract four, even we will get zero. So there are three possible number off. Positive. Real Ciro's said these therefore either four or two or none. Zero is not. Okay. So this is the possible number. A positive real zeros. No, Let us find the number off. Negative real zeros. Okay, so for that, uh, we will see here there is no change in sign, no change in sign, no violation inside. No violation. And say so. That means the number off negative real zeros are Zito. That with none no negative real zero exist. So now we will use in our second part. We will use the congressional Zito putem to find the possible zeros that this bullet email could have. So our task with this be negative X is done, so we will omit it. Okay, Now we have only in the actual political which spx Okay, fine. So no, the rational zero. Thuram says that if we take all the possible issues off the factors off this lust, ERM there is the constant, um, to the factors off the officiant of the leading term. So here the coefficient is one. So the ratio off this too will give us all the possible Ciro's. Okay, So since the denominator of the issue is one so we don't need to worry about that, we just only need to find the factors off positive 26 and both indigenous in desert factors, that means negative importante. All factors and those factors will be the possible zeros for this bulletin meal. So let us right the factors off 26 so positive, negative one than positive negative do, then, was a definitive toting and Leslie positive negative during disease. So these are the factors off 26 now, since according to the discourteous rule of sign, we have no negative zeros. So we can combine the discourteous rule off sign and the rational zero to your, um and say that the number off possible zeros are only positive. One was rift to positive 30 and positive. 26. All the negative numbers are eliminated as because there is no negative real zeros. Okay, so these are the actual possible zero's real zeros. By combining the 23 OK, no, let us test. So the third part is testing and finding order zeros. Okay, so the polynomial is this one, so I will bring it down. So let me take it over here. Okay, fine. So now we will test. It s so the 1st 1 is Ah, be off one. So that is one to the ball. Full, minus 71 cube plus 27 1 square minus 47 multiplied with one plus 26. Okay, so one to the ball for his ones than seven. I lost 27 and minus 47 plus 26. If you combine all the positive terms and the negative term separately, so it will get 26 plus 27 is ah, 53 54. Positive and negative. Seven and negative. 47. It's negative. 54 coaches equals 20 That means one is the zero off this plane email. So I will write the list of zeros on the right side. So zeros are so we got our 1st 0 which is one Okay, No, let us just for the other. Okay, I will guess for our second possible zero, which is stupid. So let us put here too. So that would be due to the power for minus seven multiplied with two cube blow Strange seven X square. That is to square minus 47 x with a stoop plus 26. So here to do the par four is 16. Okay, so to Cuba is eight and eight multiplied seven is 56. Some degree, 56 to square is four and 4 to 27 is ah, 28 years since 98. Okay, sorry. It would be 108 actually, 108. Okay, so then 47 multiplied with two is 94. Okay. And then 26. So now let us combine the positive terms. So 108 plus 16 plus 26. So we have 36 for to do who is 100? 150 and 56 80. 50 sixes and negative. 94 years. Also negative. 150. So we have zero here, So that means toe is a zero off this polynomial. So in the list, we can put too. So one. So two more zeros are left and see. Since we got two zeros, I will check another one. The last one. Okay, so we are in this case now. We can have 20 Okay, Real zero. If we find another than we will have definitely four reels. It also get so the last one would be so let us see what is our next possible zero, which just hurting so their support. 30 Nancy. So if we want to find B 13 by this conventional mattered, so it would be a lengthy process. Okay, So now I will use another method which is called Cindy Regulation method. Okay, since we know to offer zeros so we can reduce the polynomial coefficients into factors. Okay, So let us perform the Cindy Regulation with the first visit of which is one. So what are the conditions off? All the drums in the parliament Belittle mills. This one negative seven, then positive. 27. So let me let only 27 then nearly 47 Leslie, 26. So in this method we bring down the first trump and then multi play it with zero. That is we have taken one. So one multiply that with one is one. Now we will add this too. So that would be negative. Six. Then again, we with the steps. So again, one multiplied with negative sixties. Negative six. Okay, so 27 minus six is 21. Then again, one multiplied with 21 years 21 and 47 21 would be 26 negative. 26 then one multiplied with negative. 26 is negative. 26. So here we have zero. So this is the remainder. Uh, and since it is zero, that means one is our zero off the polynomial. Now we have this coefficients left. Now we can reduce again by using the seconds it'll which is to so let us divide the coffee. She, um switcher left. That is the factor of this polynomial. Is that the coefficient of the factor off this polynomial? Okay, so we will divide this sector with another Ah, Zito. Okay, it's individually. So this will give the our reduced factor again. Ok, so as usual, we bring one, then multiply with two. So then added negative four. Then again, multiply with two so that this negative eight, then multiply and then adding 21. A negative eight will give us negative 13. Sorry. Positive. 30. So 30. Okay, then two multiplied with 13. It's negative. Twenties Positive. 26. So this will add up to zero as usual. Since two is the zero, it will bring as the remainder. Okay, now we're left with this coefficients which at the provisions Ah, off the factor. Excluding the factors off one into. Okay, So the factors which has been made by one and two there's been excluded. No. Now we have only this coefficients left. So with our remaining zeros which are 13 and 26 we can not checked. So let us sick with 13. So 13 and our factor so negative one or one negative Four and 30 says you can see one will go down then we have 13 multiplied. One is 13. So adding them we will get negative nine. Okay, so 13 multiplied with negative nine is 117 So negative. 117 So here it is, not zero. It is negative. Wonderful. Okay, so this is not a goes to zero. That means 13 is not a zero off this polynomial who gets 13. It's not desert of this. Fill in a meal. So 26 will not be also the zero off this planet will. Why? Because if 10 disease becomes the zero, then 13 must also be zero. Because we have this options only four. Either 42 or zero. So we're stuck in this park, so we have to really zeros which are wanting to So these are the two rials zeros, okay. And the remaining zeros are imaginary, so these are not really imaginary numbers. Okay, so now with this really zero, we can find the factors. The last words off this problem is we have to represent political PX in terms of factors. So let me give you a device. Are revision of the factory tour So the fact of term sees there if is zero off a pole in a meal than X minus E is a factor. Okay, so since here we have the two factors, that means one and two. Okay, so we can write X minus one and x minus two are factors hoagie. So now we have to find them, find another two factors. But these two factors will not come explicitly because we know we do not have riel routes other than one and two. So that would be a quadratic equation over here in this park. And this quarter, immigration will have the coalition's the's one. So these are the coffee shops off the water. Digression, so you can write it directly from here, which is excess square. So the coalition of excess square is one than the coefficient of access nearly force in 84 X and then Leslie with 13. So later plus 30. Okay, so this is the representation of the polynomial in terms off product, off linear factors, and 1/4 defectors, which is irreducible. So this cannot be reduced to two more linear vectors. OK, so I hope even a student will thank you.

We have the question p of X because two x to the third minus X squared plus four X minus seven on. The first thing we're gonna do is we're going to use, um, the rule of signs to figure out how many, um, the zeros positive and negative there are in this equation. Um, and the rule of signs is first, you have to find the, um, variation of sign. And so, um, in this case, uh, there is this is we'll put Plus, in there there's one to three variations and sign of p of X P of X has 33 no mess as three variations and sign. And then that means here we're taking P deposit X, that there could be either three or one, um, riel, positive zeros. And that is because, um, the dig hearts rule states that you can either have the number of variation and sign of real zeros or and even whole number less. So the possibilities are three and one, um, and notifying the number of negative real zeros. We take p of negative acts, which equals, um, negative two x to the third minus X squared minus four x minus seven. And in this case, there's no variation. And zero no variation inside. And zero no operations here. I'm at no variation in sign. Um, so that means there is zero zero negative. Negative is your bows. Um and so the possible number, Actually, that's impossible. That's a different thing. The number according to Dick arts Rule signs or either three or one zeros, but the total possible number of zeros. We're gonna go back up to our, um, original polynomial up here. Come on. Sorry. Here. Um, the total possible numbers. We're going to go back to the equation. P over Q. Where P includes the factors of seven que factors of to, and so we're gonna right the total number. So P over Q. We're appeased the factors of seven que factors of to. And don't forget your poster minus. So the possible zeros are gonna be seven over to 1/2, um, one and seven, and then we're gonna dio closer minus in front of that. So if we weren't using the current rule of signs just using pure McCue, we might say that they're eight possible routes. But we know that there only could be three or one


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