# Help polynomial

1. Apr 1, 2004

### 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

2. Apr 1, 2004

### matt grime

Can you factorize x^8 - 1?

3. Apr 1, 2004

### rattis

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, im in a physics lesson.

and why do you use p(x)= ? i always thaught it was supposed to be f(x)=. although it doesnt matter

4. Apr 1, 2004

5. Apr 1, 2004

### 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.

6. Apr 1, 2004

### expscv

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 thx chen thats a good way solve it, but i was asked to solve hence by x^8-1

7. Apr 1, 2004

### Chen

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

8. Apr 1, 2004

### 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...

9. Apr 1, 2004

### 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:
$$p_{(x)} = x^6 + x^4 + x^2 + 1 = (x^2 + 1)(x^4 + 1) = 0$$
.......

10. Apr 1, 2004

### expscv

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

11. Apr 2, 2004

### rattis

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

12. Apr 2, 2004

### PRodQuanta

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.

13. Apr 3, 2004

### Muzza

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

14. Apr 3, 2004

### 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?

15. Apr 3, 2004

### expscv

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

16. Apr 3, 2004

### Chen

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

17. Apr 3, 2004

### PRodQuanta

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

18. Apr 3, 2004

### Muzza

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

19. Apr 3, 2004

### Hurkyl

Staff Emeritus
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.

20. Apr 3, 2004

Or sum a gp.