gcd in F5

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June 16th 2011, 06:52 PM
wik_chick88

gcd in F5

calculate $gcd(x^3 + 2x^2 + 3x - 1, 2x^2 - x - 1)$ in $F_{5}$

i know how to go about this question, im just a little confused...
does $6x^2 \equiv 0$ , because we are working in $F_{5}$ ?
June 16th 2011, 07:28 PM
TheEmptySet

Re: gcd in F5

Quote:

Originally Posted by wik_chick88

calculate $gcd(x^3 + 2x^2 + 3x - 1, 2x^2 - x - 1)$ in $F_{5}$

i know how to go about this question, im just a little confused...
does $6x^2 \equiv 0$ , because we are working in $F_{5}$ ?

No, if that were the case it would be the additive identity. Just reduce the coefficient mod 5

$6x^2=1x^2=x^2$
June 16th 2011, 07:47 PM
Also sprach Zarathustra

Re: gcd in F5

Quote:

Originally Posted by wik_chick88

calculate $gcd(x^3 + 2x^2 + 3x - 1, 2x^2 - x - 1)$ in $F_{5}$

i know how to go about this question, im just a little confused...
does $6x^2 \equiv 0$ , because we are working in $F_{5}$ ?

Roots of: g(x)=2x^2 - x - 1=0 are: 1 and -1/2

Say f(x)=x^3 + 2x^2 + 3x - 1,

f(1)!=f(-1/2)!=0

gcd((f(X),g(x))=1

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