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Prove that x^4+10x^3+14x^2+8x+2 is irreducible over z...

Question

Prove that x^4+10x^3+14x^2+8x+2 is irreducible over z

Prove that x^4+10x^3+14x^2+8x+2 is irreducible over z



Answers

Consider the set
$$S=\{p \in P_3 : (x+1)p'(x)-3p(x)=0 \text{for all} x \in R\}$$
1)Show that S is a subspace of P_3 ( the set of all real polynomials of degree <= 3).
2) find a polynomial in S where not all the coefficients are zero.

What we want to do is step through um to be able to find some polynomial that when this polynomial is subtracted from two X cubed minus X squared plus four eggs minus two. So I'm going to subtract this polynomial and we're just going to label it P. So when I subtract that polynomial from this polynomial, then my result is X squared plus two X minus nine. So I want to know what that P is. Okay. And so the first thing we need to know is we want positive P. So what we're gonna do is I'm going to add P to both sides. So this will be two X cubed minus X squared plus four X minus two is now equal to X word plus two X minus nine plus whatever this is, that's what we're looking for. So I want to try to get this P all by himself. So that means I'm going to have to subtract X squared from both sides. And remember I can only subtract an X squared with another X squared. I'm a. Subtract the two X from both sides and I'm going to add the nine. So when I do that the left hand side becomes two X cubed minus two X squared plus two X plus seven. And that will equal because all of these add up to zero. That is left with what my polynomial needs to be. So my polynomial needed to be to execute minus two X squared plus two X plus seven. Hope this helps.

We want to find which polynomial must be subtracted from. Two X cubed minus X square bliss. Four eggs minus two. So that the differences X cubed plus two X minus nine. So let's P. Of X. B. The polynomial to be distracted from these paranormal here. Yeah mm Two X cube minus X square plus two eggs four eggs minus two then. But he's said here is that two X cubed minus minus X squared plus four X minus two minus P of X. That is we're subtracting Pekovic from is polynomial ended and that difference must be equal to these really normal here. X cube plus two eggs minus nine mm. And now what we want to do is to find this paranormal. So we solve this equation for this polynomial P. Of X. It means we're going to put P. Of X to the right so it's going to change sign. And then this whole put normal will be best to the left. So we get to X cube minus X square plus four eggs minus two which stay on the left side of the equation, minus this whole polynomial here which we have passed to the left so minus parenthesis, X cube plus two x minus nine. It's important to notice here that the whole paranormal is passing to the left. So the sign affects all the polynomial. This negative sign here. Mhm. And then negative P of X is uh has been passed the right so the sign is going to change this P of X directly. So we have sold at once the equation to view Vicks that is P of X is equal to. We're reading from right to left now two X cube minus x square plus four x minus two. And here we distribute design negative sign inside the parenthesis. So we get negative x cube minus two X plus nine. And here we simplify the expression group in similar terms like two X cube minus X cube, that's equal to x cube. Then we have negative X square is alone, there is no other X square the equation. Soviet negative X square. Then we have four X. And negative two eggs, four x minus two eggs plus two eggs. And finally we have uh the uh independent terms or constant terms. We get nine minus 27 So the polynomial that must be subtracted from two X cubed minus X squared plus four x minus two. To get as a result X cubed plus two X minus minus nine is X cubed minus X square plus two X plus seven. Yeah so this is the polynomial that must be subtracted. Yeah. Mhm. For um two X cubed minus x square plus for x minus two. So that difference yeah difference. Okay. Right. Mhm. Is X cube plus two eggs minus night. So this is the answer to the brook. Uh huh. Uh huh.

Hello. Then we have a problem. X ray to about four plus two X. Q plus 22 weeks Squire plus 50 X minus 75 equal to zero. And we must have to find the we have to find the zeros of this polymer. Or we have to solve this equation. Okay, so let's see. You hit general method first. So if you plug in X equal to one, we will be having one plus two plus 22 plus 50 minus 75. We call to zero. So this is 75 minus 75. We called +20 Which is true. So X equals one is one of the zeros. So X minus one is one of the factors. Now if we all just try and plug in minus one, let's see what we got. Let us see ah one 22 plus 1 23 23 23 23 minus 2 21. 21 minus 50. No. Uh huh. Okay. So let us just divide this whole equation with x minus one to get another zeros. So exodus depart for bless to excuse. Plus 22 X esquire plus 50 X minus 75 here x minus one. Okay so let us write xQ here X cubed minus xQ excited about four minus excuse. Sorry, expressed the powerful minus X q minus plus. It cancelled out. So three X q plus 22 access choir. So let us write three access query so three xq minus three X squared minus place. So 25 X squared plus 50 X. So plus 25 X here 25 X squared minus 25 X. That is 75 X minus 75. So here plus 75 you should. Right so 75 x minus 75 minus plus get counselor. So we have yeah x minus one in two. Okay so this is a question to execute projects, Squire X cubed plus three X esquire. Okay Plus 25 X plus 75 X cube plus three X squared plus 25 X plus 75 Equal to zero. Yeah, this is the uh factories form. No, we have to solve this equation. Okay, this is cubic equation. So if you plug in X equal to minus three in this equation we're talking about X equal to minus three will be getting minus 27 minus and plus 18 minus 75 plus 75. Oh okay, okay this will be 27 Sorry, 27. So this becomes equal to zero. So we have X equal to minus three as one of the factors X equal to minus three is one of the factor of this cubic equations. So let us just start doing the same thing that is by dividing exit. Uh This equation by X plus three. That is cubic equation by X plus three X cubed plus three X squared plus 25 X plus 75 divide by X plus three. So this is X squared x q plus three X squared minus minus 25 X plus 75 plus 25 25 X plus 75 minus minus zero. So now the whole expression that is I'm talking about this expression or equation because right hand side is zero, will build an X minus one in two explains three in two, X squared plus 25 to 0. So we have zero X equal to one, X equal to minus three and access square plus 25 equal to zero. So access square will be equal to minus 25 which will give us imaginary roots. So five I hold square X will be equal to plus minus firefighter. So there are only two year roots and two imaginary roots, and the total number of routes. Real roots is two railroads are two X equal to one and minus three, and including even the route will be having minus five way and plus five by. Did that answer?

Mhm. We are given a space as that contains polynomial of degree three year. Lower that fulfills a given equivalence Part one asked us to show that this set S. Is a subspace of Polynesia polynomial of degree three. And the solution is quite straightforward since within the definition of the set it only takes polynomial polynomial from Polynomial of degree three. The birth, so all all polynomial and S. They will be in polynomial three. A Parliament of decrease three. Mhm. Uh There does exist the polynomial in S. It's a non empty space as it as we have a table here showing different coefficients for for the different powers of X. In the equation. X cubed plus B, X squared plus C. X plus deep. Okay. Uh so there exists upon them inset therefore some empty set. Therefore it's a proper subspace. And we are done moving on. Part two. We are asked to find a polynomial in S where not all the coefficients are zero. So we opened up a class for all polynomial is in X. We can denote can denote the polynomial as such. Okay. Yeah. Uh I have also noted three times the polo meal here and then the polynomial different differentiated I assume uh listener knows how to differentiate polynomial and I'm going to move on and then I multiplied the differentiated polynomial by X plus one giving us this formula. And finally I am using the equation that was given to us in the definition of set S. Yeah. So we can drive several things from this equation over here. Uh We need this needs to hold true for all access so we need three A minus B uh equals zero. Or in other words we need 38 equals B. Since two B -2 C need to be zero. Include that the equal. See And since it contains three And also contains C -3 d. You conclude that C Must be equal to three D. So you can just pick A # 40. Like one seems to be three times team. So it's three. He needs to be seen. So It's the same as seen was three. And there needs to be three times be. So it's nine. I did the same thing for D equals two. And he goes three. You see that this is true for any real number I just the D equals I. C. Equals three, I. B equals three I. And equals nine i. So our space s can be rewritten as such with it. You can derive any uh any polynomial from s in a very simple manner.


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