Factorization of Polynomials over a field

  • #1
I don't understand how to factor a polynomial over Z3 [x], Z7 [x], and Z11 [x]

I need to factor the polynomail x3 - 23x2 - 97x + 291

PLEASE HELP!!!
 
Last edited:

Answers and Replies

  • #2
Note that for a polynomial is of degree 2 and 3, reducibility is equivalent to the existence of roots.

mod 3:

X^3-23X^2-97X+291=X^3+X^2+2X=X(X^2+X+2). A calculation shows X^2+X+2 doesn't have a root in Z/3Z. Done.

mod 7:
X^3+5X^2+2X+4. A calculation shows it has no root in Z/7Z. The polynomial is irreducible.

mod 11:
291=5=1*5=10*6. So if the polynomial has a root, it should be 1, 5, 6, or 10. A calculation shows X^3-X^2+2X+5 has no root. The given poly is irreducible.
 
  • #3
I think I might have phrased this question wrong, but I figured it out.

THANKS
 

Related Threads on Factorization of Polynomials over a field

Replies
1
Views
5K
Replies
2
Views
775
Replies
15
Views
13K
Replies
1
Views
2K
Replies
5
Views
2K
Replies
5
Views
2K
  • Last Post
Replies
8
Views
3K
Replies
1
Views
2K
Replies
5
Views
3K
Top