Irreducible factorization of polynomials?

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ricsi046

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Hello
Can you tell me how can you give the irreducible facorization of polynomials over R,Q and Z5 (or some other prime number instead of 5)?
I have a few example:
x6-125 in R
x 6 + 3x5+6 x4+14x 3+16x2+10x+4 in Q
x 5 + 2x4+3 x2+4 in Z5
The first one is simple,i just find the roots,taking the 6th root of 125.In the 2nd i can use the following theorem:Let f=a n xn+...+a 1 x+a0 and gcd(p,g)=1.If f(p/q) =0,then q|a n p|a0 .So we get a finite set of the possible roots.

What about Z5 ?Is it correct what i wrote?
 
ricsi046 said:
Hello
Can you tell me how can you give the irreducible facorization of polynomials over R,Q and Z5 (or some other prime number instead of 5)?
I have a few example:
x6-125 in R
x 6 + 3x5+6 x4+14x 3+16x2+10x+4 in Q
x 5 + 2x4+3 x2+4 in Z5
The first one is simple,i just find the roots,taking the 6th root of 125.In the 2nd i can use the following theorem:Let f=a n xn+...+a 1 x+a0 and gcd(p,g)=1.If f(p/q) =0,then q|a n p|a0 .So we get a finite set of the possible roots.

What about Z5 ?Is it correct what i wrote?
Click to expand...
The "numbers" in \(\displaystyle Z_5\) are 0, 1, 2, 3, and 4. There is no "5" so cannot be a \(\displaystyle x^5\). In \(\displaystyle Z_5\), \(\displaystyle x^5= x^0= 1\). That should be just \(\displaystyle 1+ 2z^4+ 3x^2+ 4= 2z^4+ 3x^2= x^2(2x^2+ 3)\). And it is easy to see that \(\displaystyle 2(1^2)+ 3= 0\) and \(\displaystyle 2(4^2)+ 3= 0\) in \(\displaystyle Z_5\) so we have \(\displaystyle x^2(x- 1)(x- 4)\).

Yes, finding the 6th roots of 125 is a good start but only two of those are real so you will have two linear factors and an irreducible fourth degree factor.

Yes, you can use that theorem to search for rational roots. What did you get?
 
Yes, those are possible rational roots. But what are the rational roots?

(It should be obvious that no positive number is a root. Which of -1, -2, and -4 actually are roots.)
 
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