Is x^4 - 14x^2 + 9 = 0 irreducible in Q?

[Machinus]

How can I prove that x^4 - 14x^2 + 9 = 0 is irreducible in Q? When I went to check quadratics in mod5 I get a lot...do I have to do long division on all of those?

 

[shmoe]

If x^4-14x^2+9 has no zeros mod 5 you don't have to check any of the quadratics which do have roots.

There is another approach if you find the mod p test keeps failing. You can find all the real roots without too much trouble. If x^4-14x^2+9 factored into quadratics over the rationals, then the roots of these quadratics would have to be chosen from the real roots. This very much limits the possibilities for the quadratics, you can check them all to see if they are in Q[x].

 

[mathwonk]

well if there were a linear factor, there would be a rational root, hence of form a/b where a is a factor of 9. so checkm that 1, -1, 3,-3, 9,-9 are not roots.

then if not, there are not linear or cubic factors. so consider quadratic factors. but if it had quadratic factors, either thye involve x or not. if not then you can use the quadratic formula toi find them, setting u = x^2. if they do involve x, then the x's are lost in the product so the two constant terms are equal and opposite in sign.

ah phooey, i get bored with these type of problems. they are too tedious.

i.e. i am not very good at algebra.