complex no. question no. 8

• May 22nd 2008, 11:26 AM
afeasfaerw23231233
complex 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 AM
topsquark
Quote:

Originally Posted by afeasfaerw23231233
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.

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 AM
Moo
Quote:

Originally Posted by topsquark
You mean aside from it not being true?

$\displaystyle x^5 - 1 = (x - 1)(x^4 +{\color{red}x^3}+ x^2+{\color{red}x} + 1)$

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

:eek:

(good luck with your boy :p)
• May 22nd 2008, 11:40 AM
Moo
Hello,

Quote:

Originally Posted by afeasfaerw23231233
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.

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

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 PM
mr fantastic
Quote:

Originally Posted by afeasfaerw23231233
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.

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 PM
topsquark
Quote:

Originally Posted by topsquark
You mean aside from it not being true?

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

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

Like I said!

Okay, I goofed. Thanks for the catch, Moo. :)

-Dan