Thread: GCD of Polynomial

I found the GCD by the Euclidean Algorithm of
(x^5)+(x^4)+(2x^3)-(x^2)-x-2
AND
(x^4)+(2x^3)+(5x^2)+4x+4
to be (4x^2)+4x+8.

But my book says something about it having to be monic. How would I manipulate (4x^2)+4x+8 to make it the true GCD of the two polynomials?

Originally Posted by JaysFan31

I found the GCD by the Euclidean Algorithm of
(x^5)+(x^4)+(2x^3)-(x^2)-x-2
AND
(x^4)+(2x^3)+(5x^2)+4x+4
to be (4x^2)+4x+8.

But my book says something about it having to be monic. How would I manipulate (4x^2)+4x+8 to make it the true GCD of the two polynomials?

My book says nothing about the GCD having to be monic, but if you would like your polynomial to be monic, then its leading coefficient is 1. So divide your polynomial by 4:
is your monic GCD.

NOTE: If you require your GCD to be monic then the GCD is unique. This may be part of your book's reasoning.

-Dan

Thanks.

Ok, similar problem.
GCD of (x^4)+(3x^3)+2x+4 and (x^2)-1 in Z5[x].
I did the Euclidean Algorithm by dividing the polynomials. The first remainder was 5x+5. Then 5x+5 divides (x^2)-1 evenly so that the remainder is zero. Shouldn't the gcd be 5x+5 and then monic x+1? My book says (x^2)-1 is the gcd. Also how does the Z5[x] change it from just Q[x]?

Originally Posted by JaysFan31

Thanks.

Ok, similar problem.
GCD of (x^4)+(3x^3)+2x+4 and (x^2)-1 in Z5[x].
I did the Euclidean Algorithm by dividing the polynomials. The first remainder was 5x+5. Then 5x+5 divides (x^2)-1 evenly so that the remainder is zero. Shouldn't the gcd be 5x+5 and then monic x+1? My book says (x^2)-1 is the gcd. Also how does the Z5[x] change it from just Q[x]?

What is 5x+5 in Z5[x]? It's zero. Thus divides and thus is the GCD.

-Dan

Hello, JaysFan31!

I found the GCD by the Euclidean Algorithm of:
and to be: . . . This is not true

Factor: .

. . .


Factor: .


Therefore: .