polynomials [another one]

• Dec 17th 2009, 09:27 AM
metlx
polynomials [another one]
You have 2 polynomials

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

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

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

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

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?
• Dec 17th 2009, 09:55 AM
metlx
that's what i tried:

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

I need to find r(x) ->

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

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

I continue with algorthm:

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

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

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

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

Continuing with algorithm:

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

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

I need to find $r_2(x)$

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

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?
• Dec 17th 2009, 10:47 AM
Raoh
here's what i think a hint(Rofl),
$x^4+1=x(x^3-1)+(x+1).$
$x^3-1=(x+1)(x^2-x+1)-2.$
• Dec 17th 2009, 11:28 AM
metlx
oops (Blush)
i feel stupid now. :P

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

any hints for b)? :P