let-yfx-be-a-quartic-polynomial-with-leading-coefficient-a1-and-fif2-i0-write-an-equation-for-f

Name: let yfx be a quartic polynomial with leading coefficient a1 and fif2 i0 write an equation for f
Uploaded: 2021-07-12
Duration: 2 min 40 s
Channel: Numerade
Description: Let $y=f(x)$ be a quartic polynomial with leading coefficient $a=1$ and $f(i)=f(2 i)=0 .$ Write an equation for $f .$

Question

Let $y=f(x)$ be a quartic polynomial with leading coefficient $a=1$ and $f(i)=f(2 i)=0 .$ Write an equation for $f .$

Kian Manafi · Accepted Answer

So we&#x27;re told that F of I. Is equal to F two I, which is equal to zero. And we&#x27;re also told that this is a core tech polynomial that we want to find with a leading coefficient is equal to one. So for this to be a quartet polynomial, we can have up to four zeros. And if we want this to be a real valued or have real coefficients, this cortex polynomial, then we&#x27;re gonna have to use the conjugate um complex zeros theorem, which tells us if we have a complex zero, which we can see that I and two ir zero for this function since F of X is equal to zero at those two X values. Um And if we have one complex um zero, then we also need to have the conjugate of that complex zero to have real valued coefficients for our core tech polynomial. So that would mean that F of negative I. Is equal to F of negative two I which is equal to zero. And so now we can actually set up this polynomial by um making it a product of factors where we have x minus I X plus I X minus two, I an x plus two I. And then let&#x27;s multiply these together. So let&#x27;s multiply these first two. We&#x27;re going to get X squared and then we&#x27;re going to get minus I squared. Since we&#x27;re gonna get negative I. X. And positive I X. So those are going to cancel. So this is going to be X squared um minus negative or the square of negative one squared plus one. And then this is multiplied by X times x minus two, I times X plus two I is going to be X squared. And then we&#x27;re going to have minus for I squared. So this is equal to X squared plus one, multiplied by x squared plus four. And so now we can do this last multiplication here. So we&#x27;re gonna get X to the fourth, going to get plus four X squared, and then we&#x27;re going to get plus X squared and plus four. So this is equal to X the fourth Plus five x squared plus four. And we can see that our lives are leading coefficient A here was supposed to equal one, and we can see that it is equal to one here, since we just have this X to the fourth. So this is a core check polynomial with the given conditions on it um that we found just by using the complex con gets zeros there.

Heather Zimmers · Answer

in this problem, we&#x27;re going to find the equation of a cubic polynomial. We know a couple of it zeros. We know it has a zero at two and we know it has a zero. It I And if it has a zero, it I It also has a zero at negative I the congregate. So the zero&#x27;s air to I a negative I So we can write this out as X equals two X equals I and X equals negative I. And then if we subtract two from both sides of that equation, subtract I from both sides of the next equation and add I to both sides of the third equation. We get our linear factors X minus two times X minus I times x plus I Let&#x27;s multiply X minus I and X plus I together. Since those Air Contra gets the insides and outsides cancel on, we&#x27;re left with just X squared minus. I squared and remember that I squared is negative one, So that factor is actually just X squared, plus one. Let&#x27;s multiply X squared plus one by X minus two. Using the oil method when we get X cubed minus two X squared plus X minus two. So that could be our function f of X. But there&#x27;s one more detail. Remember, it said in the problem that the leading coefficient plus to be negative one so right now are leading coefficient is positive one. So we need to multiply this whole equation by negative one for this whole expression, I should say by negative one, and that will give us our function f of X. So the opposite of X cubed plus two x squared minus X plus two and that meets all the requirements.

Heather Zimmers · Answer

in this problem. We&#x27;re looking for the equation of a Kordic polynomial, so that means it&#x27;s fourth degree. The leading coefficient is one, and we also know that after by zero and f of two, I equals zero. That means we know we have zeros at I and two, I remember that if you have a non really zero, it&#x27;s conjugated is also a zero. So that means we also have negative I and negative to I. So now we know the four zeros, and what we can do for each zero is right. A linear factor. So if X equals I, it&#x27;s linear factor would be X minus. I we have X equals two I. It&#x27;s linear factor would be X minus two. I, we have X equals negative I It&#x27;s linear factor would be X plus I. Now we have X equals negative two y. It&#x27;s linear factor with the X Plus two way. Now let&#x27;s take those, rearrange the order and multiply them together to get our Kordic polynomial. So I&#x27;m going to multiply X minus. I and X minus are seeing the X plus I since those air congregates they multiply together fairly easily, and I&#x27;m going to multiply X minus two I and X plus two I Same thing. We have congregates. Those will multiply together nicely. So we multiply the 1st 2 using the oil technique and we get X squared. Plus I x minus I x minus y squared So the insides and outsides cancel. We multiply the last ones together and we get X squared plus two y x minus two y x. So the insides and outsides cancel minus four. I squared and remember, I squared is negative one. So we actually have X squared, plus one times X squared, plus four. So let&#x27;s use the foil techniques to multiply these together we have X to the fourth plus four x squared plus one x squared, plus four. So this gives us next to the fourth plus five x squared plus four. It does have a leading coefficient of one and this is our function F of X

Ankit Gupta · Answer

we have given a cushion. Why is equal to FX? Be 1/4 degree polynomial with a leading coefficient is equal to minus one and plus minus one is equal toe bless. Minus two is equal to zero. The desire zero effects. They don&#x27;t dysfunction. So what we need to do? We need to find FX. So we have given this. So if yeah bless minus one is in Quito air bless. Minus toe is equal to zero. Then weaken Simply guided factor within the corridor X plus minus for we will use plus one and another one X minus one. And for this we can diet X plus two and x minus. And here coefficient is written minus one So we can use minus one here, and this will become functional effects. So we need to solve this. Get done. So So when we are solving this is this is the formula formula, former love it is. Is there minus this card with equal toe a plus B and a minus. So yes, a pres de any minority. So we can simply write this like this excess Glad minus one and see. Same With this, we can use excess crap Way. This And here it is a minus. So after calculating, we will get next. Full bless five X is square minus folks, but this is a value fix.

Kristen Karbon · Answer

and this problem were asked to find a polynomial that has a leading coefficient of one that has the following routes and multiple cities. So starting with each one writing it as a factor. If the route is to then the factor is X minus two multiple cities, one in all of these. So they&#x27;re all raised to the first power. And then if the route is negative two than we have X minus negative to which of the X Plus two Oh X minus two I and then X plus two I Once I have the factors, then I can multiply this out. So taking two at a time, this would be X squared minus four and multiply needs to. Together I would get X squared minus four. I squared, which is the same thing as plus four. Now multiplying those out. I&#x27;ll get X to the fourth, but to inside terms cancel and then I get minus 16. So half of axe equals that polynomial

Kristen Karbon · Answer

were asked to write a polynomial with a leading coefficient of one that has the following routes and multiple cities. So starting here, if I have a root of zero and a multiplicity of two, that means my factors X and the multiplicity of to meet its X squared. And then on the next one, If my route iss four, that means I have a factor of X minus four. And the multiplicity is one right. Once we have our factors, then we&#x27;re going to change this by multiplying it out. So this is gonna be X to the third minus four X squared, which can be written as f of X, equals that polynomial.

Problem

Let $y=f(x)$ be a cubic polynomial with leading c…

Problem 133 Easy Difficulty

Let $y=f(x)$ be a quartic polynomial with leading coefficient $a=1$ and $f(i)=f(2 i)=0 .$ Write an equation for $f .$

Answer

$f(x)=x^{4}+5 x^{2}+4$

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$f(x)=x^{4}+5 x^{2}+4$

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Ron Larson, David C. FalvoPrecalculus with Limits

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