(New page: Are the following irreducible over Q? a) <math>x^5 + 9x^4 + 12x^2 + 6</math> b) <math>x^4 + x + 1</math> c) <math>x^4 + 3x^2 + 3</math> d) <math>x^5 + 5x^2 + 1</math> e) <math>(5/2)x^5 + ...)
 
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d) <math>x^5 + 5x^2 + 1</math>
 
d) <math>x^5 + 5x^2 + 1</math>
 
e) <math>(5/2)x^5 + (9/2)x^4 + 15x^3 + (3/7)x^2 + 6x + (3/14)</math>
 
e) <math>(5/2)x^5 + (9/2)x^4 + 15x^3 + (3/7)x^2 + 6x + (3/14)</math>
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a.) Look at Eisenstein's with p = 3.
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b.) A polynomial is irreducible in Q if there's a p such that f(x) mod p is irreducible.  Look at p = 2.
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c.) See part a.
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d.) See part b.
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e.) Multiply by 14 then see part a.
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--[[User:Jniederh|Jniederh]] 22:12, 8 April 2009 (UTC)

Revision as of 17:12, 8 April 2009

Are the following irreducible over Q?

a) $ x^5 + 9x^4 + 12x^2 + 6 $ b) $ x^4 + x + 1 $ c) $ x^4 + 3x^2 + 3 $ d) $ x^5 + 5x^2 + 1 $ e) $ (5/2)x^5 + (9/2)x^4 + 15x^3 + (3/7)x^2 + 6x + (3/14) $



a.) Look at Eisenstein's with p = 3. b.) A polynomial is irreducible in Q if there's a p such that f(x) mod p is irreducible. Look at p = 2. c.) See part a. d.) See part b. e.) Multiply by 14 then see part a. --Jniederh 22:12, 8 April 2009 (UTC)

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Ryne Rayburn