Essentially my question is to find the GCD of the following polynomials using the Euclidean algorithm:
$A:= x^5-3x^4+3x^3-2x^2+2x-4$
$B:= x^5-x^4-3x^3-x^2+x+3$
My attempt:
(Each new remainder will be the following letter of the alphabet. )
$A = 1B +(-2x^4+6x^3-x^2-3x-7) \\ B = (-\frac{x}{2}-1)C+(\frac{5}{2}x^3-\frac{7}{2}x^2-11\frac{x}{2}-4) \\C=(-\frac{4}{5}x+\frac{32}{25})D + (-\frac{23}{25}x^2+\frac{21}{25}x-\frac{47}{25}) \\ D=(-\frac{125}{46}x+\frac{700}{529})E \ + (-\frac{6200}{529}x-\frac{800}{529} \\ E= (\frac{12167}{155000}x - \frac{393047}{4805000})F+(-2.0037)$
At this point I assumed the GCD is $1$ as the remainder is constant.. however I'm 99% sure this is wrong due to the actual factorisation of A and B.
$ A = (x-2)(x^2-2x+2)(x^2+x+1) \\ B = (x-1)(x^2-x-3)(x^2+x+1)$
They clearly have a common factor so I'd expect that $GCD(A,B) = (x^2+x+1) $
I'm assuming I'm misunderstanding something fundamental in terms of the Euclidean Algorithm but not entirely sure what.. any help is appreciated!
$C = A - B = -2x^4 + 6x^3 - x^2 + x - 7\\ D = B + (\frac 12 x + 1)C = \frac 52 x^3 - \frac 32 x^2 - \frac 32 x - 4$
This appears to be your mistake.
There is no reason why you can't say.
$D = 2B + (x + 2)C = 5 x^3 - 3x^2 - 3x - 8$ to keep it all with integer coefficients.
And sometimes half steps are easier to keep track of.
$E = 5C + 2Dx = 24x^3 - 11x^2 - 11x - 35\\ F = 24D - 5 E = 17 x^2 +17 x + 17\\ G = x^2 + x +1$