Solving Polynomial: p(x) = x^6 + x^4 + x^2 + 1 | Factorization

[expscv]

if p(x ) = x^6 + x^4 + x^2 + 1

show that the solutions of the equation p(x ) = 0 are among the solutions of the equation x^8 - 1 = 0

hence factorse p(x ) fully over R

 

[matt grime]

Can you factorize x^8 - 1?

 

[rattis]

Can you factorize x^8 - 1?

no, but you can factorise p(x ) = x^6 + x^4 + x^2 + 1.

although i have not got time and resources to do it, I am in a physics lesson.

and why do you use p(x)= ? i always thaught it was supposed to be f(x)=. although it doesn't matter

 

[Chen]

https://www.physicsforums.com/showpost.php?p=174426&postcount=5

 

[matt grime]

well, you can factorize x^8-1, and it's because you can do that easily that you can factorize p(x) {p is just a label, it doesn' mean anything, Rattis, if you were taught always to use f how did you cope if there was more than one object to label?}. You don't need any (extra special) resources, you can do it in your head.

 

[expscv]

Can you factorize x^8 - 1?

yeah i think so x^8-1 = (x-1)(x^6+x^5+x^4+x^3+x^2+x+1)

but it seems useless



and thanks chen that's a good way solve it, but i was asked to solve hence by x^8-1

 

[Chen]

You can factorize x^8 - 1 easily using a^2 - b^2 = (a + b)(a - b) repeatedly.

 

[matt grime]

But you can factorize it more than that.

Hint

let y=x^2, factorize y^4 - 1 as (y-1)(1+y+y^2+y^3), which you know. now put the x back in, what do you get? This is just saying that 1 and -1 are roots of x^8-1, and we factor them out to get...

 

[Chen]

It's just:
x^8 - 1 = (x^4 - 1)(x^4 + 1) = (x^2 - 1)(x^2 + 1)(x^4 + 1) = 0
And since p(x) can be factorized to:
p_{(x)} = x^6 + x^4 + x^2 + 1 = (x^2 + 1)(x^4 + 1) = 0
...

 

[expscv]

wow surpised how came i never thinked abt that? ~~~

 

[rattis]

well, you can factorize x^8-1, and it's because you can do that easily that you can factorize p(x) {p is just a label, it doesn' mean anything, Rattis, if you were taught always to use f how did you cope if there was more than one object to label?}. You don't need any (extra special) resources, you can do it in your head.

resources as in time, and the fact that the teacher dosent notice me

 

[PRodQuanta]

rattis said: and why do you use p(x)= ? i always thaught it was supposed to be f(x)=. although it doesn't matter

matt grime said:. . .p(x) {p is just a label, it doesn' mean anything, Rattis, if you were taught always to use f how did you cope if there was more than one object to label?}.

Actually, it is f(x), where f stands for function and the number inside your parenthesis (or x in this case) is the variable that the function is operating on.

Paden Roder

 

[Muzza]

PRodQuanta, are you saying that it's wrong to use other letters than f to indicate a function...?

 

[matt grime]

So, I've gto f(x) =sin(x), and I want another function to equal cos(x), and I must call that f as well must I?

 

[expscv]

doesnt matter i guess, just incdicate group/function 01 =f(x) group/function 02 = g(x)

 

[Chen]

Actually, it is f(x), where f stands for function and the number inside your parenthesis (or x in this case) is the variable that the function is operating on.

Can you please define the derivative of the product of two different functions using f(x) alone?

 

[PRodQuanta]

Hey, that's just what I have been taught. It must be wrong then. Sorry.

Paden Roder

 

[Muzza]

Yeah, it's pretty wrong, considering there are functions like sin(x), det(A), log(x) and *gasp* \phi(n). ;)

 

[Hurkyl]

If you don't see the sneaky way to factor it, the most straightforward approach is to just compute the 8 roots of x⁸-1 and then plug them all into 1 + x² + x⁴ + x⁶.

 

[matt grime]

Or sum a gp.