1. ## polynomials [another one]

You have 2 polynomials

$\displaystyle p(x) = x^4 + 1$
$\displaystyle q(x) = x^3 - 1$

$\displaystyle d(x)$ is their greatest common divisor.

a) Find $\displaystyle d(x)$
b) Find such $\displaystyle f(x)$ and $\displaystyle g(x)$ so that the following is true:

$\displaystyle p(x)f(x) + q(x)g(x) = d(x)$

For a) i tried with Euclid's algorithm but it didn't give me any good results (or maybe i did it wrong,, is it done differently than with numbers?)

no clue for b).

can anyone help?

2. that's what i tried:

$\displaystyle (x^4 + 1) = a \cdot (x^3 - 1) + r(x)$

I need to find r(x) ->

$\displaystyle \frac{(x^4 + 1)}{x^3 - 1} = x + \frac{1 + x}{x^3 - 1}$

a = x
$\displaystyle \frac{1 + x}{x^3 - 1} = r(x)$

I continue with algorthm:

$\displaystyle x^3 - 1 = b \cdot \frac{1 + x}{x^3 - 1} + r_1(x)$

I need to find $\displaystyle r_1(x)$ ->

$\displaystyle \frac{x^3 - 1}{\frac{1 + x}{x^3 - 1}} = \frac{x^6 - 2x^3 + 1}{x +1} = x^5 - x^4 + x^3 - x^2 + x - 1 + \frac{2}{x + 1}$

$\displaystyle \frac{2}{x + 1} = r_1(x)$
$\displaystyle x^5 - x^4 + x^3 - x^2 + x - 1 = b$

Continuing with algorithm:

$\displaystyle \frac{1 + x}{x^3 - 1} = c \cdot r_1(x) + r_2(x) \Rightarrow$

$\displaystyle \frac{1 + x}{x^3 - 1} = c \cdot \frac{2}{x + 1} + r_2(x) =$

I need to find $\displaystyle r_2(x)$

$\displaystyle \frac{\frac{1 + x}{x^3 - 1}}{\frac{2}{x + 1}} = \frac{x^2 + 2x + 1}{2x^3 - 2} = c + r_3(x)$

But that's where i get confused by it.. the Power of the first one is smaller than second one -> can't divide..

What am i doing wrong?

3. here's what i think a hint,
$\displaystyle x^4+1=x(x^3-1)+(x+1).$
$\displaystyle x^3-1=(x+1)(x^2-x+1)-2.$

4. oops
i feel stupid now. :P

so the greatest common divisor is $\displaystyle - \frac{x^2 - x +1}{2}$ then?

any hints for b)? :P