p307 q10

prove that

$\displaystyle x^5 -1 = (x-1)(x^2-2x cos \frac {2\pi}{5} +1)(x^2-2xcos\frac{4\pi}5+1)$

would anyone just give me some hints on this type of question? thanks.

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- May 22nd 2008, 11:26 AMafeasfaerw23231233complex no. question no. 8
p307 q10

prove that

$\displaystyle x^5 -1 = (x-1)(x^2-2x cos \frac {2\pi}{5} +1)(x^2-2xcos\frac{4\pi}5+1)$

would anyone just give me some hints on this type of question? thanks. - May 22nd 2008, 11:32 AMtopsquark
You mean aside from it not being true?

$\displaystyle x^5 - 1 = (x - 1)(x^4 + x^2 + 1)$

I assume since the title of the post involves complex numbers that you left two i's out of this. (The kid is crying so I don't have time to figure out where. :) )

-Dan - May 22nd 2008, 11:39 AMMoo
- May 22nd 2008, 11:40 AMMoo
Hello,

$\displaystyle x^5-1=(x-1)(1+x+x^2+x^3+x^4)$

A polynomial of degree 4 is necessarily a product of two polynomials of degree 2.

Why ?

This has to do with the d'Alembert-Gauss theorem, which states that any polynomial can be developped into a product of polynomial of degree 1 containing complex numbers.

However, I don't know (or don't remember I've learnt it..) why a polynomial can always be a product of polynomials of degree 1 or 2... (Worried)

Getting back to the problem :

$\displaystyle x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)$

Develop in order to find a, b, c and d...

Edit : it seems that the first attempts are vain for solving.. - May 22nd 2008, 04:10 PMmr fantastic
The attachment I gave at this thread: http://www.mathhelpforum.com/math-he...-hard-way.html

might be of indirect interest to you. - May 22nd 2008, 07:28 PMtopsquark