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$f=x^4-x^3+14x^2+5x+16$,

  1. considering it a polynomial with coefficient in $\mathbb{F}_3$, it has no roots

  2. Considering it a polynomial with coefficient in $\mathbb{F}_3$,it is a product of two irreducible factor of deg $2$ over $\mathbb{F}_3$

  3. Considering it a polynomial with coefficient in $\mathbb{F}_7$,it has a irreducible factor of deg $3$ over $\mathbb{F}_7$

  4. it is a product of two poly of deg $2$ over $\mathbb{Z}$.

1 is true, I checked. For 2, I am not sure after simplification I came to the expression $x^4+x^3+2x^2-2x+1$, but how to conclude then?For 3 and 4 ,I do not know how to do. please help.

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1  
Alpha is your friend: "factor x^4-x^3+14*x^2+5*x+16 modulo 7". (4) is obviously wrong: were it true, there would be a factor of degree 2 in (3), just take factorization over $\mathbb{Z}$, then modulo 7. –  Jean-Claude Arbaut Mar 15 '13 at 11:08
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For 3, if it has an irreducible factor of degree $3$, what is the degree of the other factor? Can you find this factor using $7$ polynomial evaluations? –  Dilip Sarwate Mar 15 '13 at 11:11
    
Another hint for part 3: Modulo 7 we have $$f=x^4-x^3-2x+2.$$ The signs alternate suggestively. –  Jyrki Lahtonen Mar 15 '13 at 13:02

1 Answer 1

up vote 0 down vote accepted

For 2, we have $$x^4+x^3+2x^2-2x+1=(x^4+x^3+x^2)+(x^2-2x+1)=x^2(x^2+x+1)+(x^2+x+1)=(x^2+1)(x^2+x+1).$$You can try to do something similar for 3 and 4.

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