Is x^4+x+1 irreducible in Q? I've tried using the einstien criterion with doing a substitution for x, but that is getting me nowhere. I also tried to factor it into two general quadratics of the form (x^2+ax+b)(x^2+cx+d) but this isn't getting me anywhere either. I konw there are no linear or cubic factors, I just need to see if there are any quadratic factors.
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Use Gauss's Lemma which states that a polynomial in Z[x] is irreducible in Z[x] if and only if it is irreducible over Q[x] and primitive over Z[x].(the statement is in fact more general in Unique Factorization Domains).
All you need to do is test irreducibility in Z[x]; you can try and construct a factorization and solve for the coefficients (which you tried already, but at least now you only need to check integer values).
(x^2+ax+b)(x^2+cx+d) yields the following equations
a+c=0
ac+b+d=0
bc+da=1
bd=1
Since these are all integers, we immediately have that b=d=1 or b=d=-1; first replace c by -a (using the first equation). Then
-a^2 + b+d=0
-ba+da=0
Note that the second of these is redundant because d-b=b-d=0, whatever they may be. So you are left with just
-a^2 + b+d=0
Just try both cases: if b=d=-1 the a^2=-2, which makes a complex. If b=d=1 then a^2=2, makes a irrational. Then there is no solution to this system and a factorization over Z[x] does not occur.
Seeing that the polynomial doesn't factor as a linear and a cubic is pretty straightforward (as you said, you've already done it). Simply check if there are roots (either by checking with some calculus or reducing the polynomial modulo(p) for some prime p; in fact any prime will work in this case, and mod2 is the easiest).
Then the polynomial is irreducible over Z[x] and by Gauss must be irreducible over Q[x].
All you need to do is test irreducibility in Z[x]; you can try and construct a factorization and solve for the coefficients (which you tried already, but at least now you only need to check integer values).
(x^2+ax+b)(x^2+cx+d) yields the following equations
a+c=0
ac+b+d=0
bc+da=1
bd=1
Since these are all integers, we immediately have that b=d=1 or b=d=-1; first replace c by -a (using the first equation). Then
-a^2 + b+d=0
-ba+da=0
Note that the second of these is redundant because d-b=b-d=0, whatever they may be. So you are left with just
-a^2 + b+d=0
Just try both cases: if b=d=-1 the a^2=-2, which makes a complex. If b=d=1 then a^2=2, makes a irrational. Then there is no solution to this system and a factorization over Z[x] does not occur.
Seeing that the polynomial doesn't factor as a linear and a cubic is pretty straightforward (as you said, you've already done it). Simply check if there are roots (either by checking with some calculus or reducing the polynomial modulo(p) for some prime p; in fact any prime will work in this case, and mod2 is the easiest).
Then the polynomial is irreducible over Z[x] and by Gauss must be irreducible over Q[x].
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if x > - 1 , then x+1 > 0, and since x^4 > 0, then p(x) = x^4 + x + 1 > 0
if x <= -1 < 0, then x^2 = x*x >= (-1)*x = -x > 0, so x^4 >= - x
so x^4 + x >= 0, and p(x) >= 1 > 0
so for any real x, p(x) is never 0
p(x) is irreducible in R (and therefore in Q).
(Note: one could have also used the rational root theorem for irreducibility in Q only.)
/Edit: Cripes! yes, sorry, was too quick. Thumbs down well deserved!
if x <= -1 < 0, then x^2 = x*x >= (-1)*x = -x > 0, so x^4 >= - x
so x^4 + x >= 0, and p(x) >= 1 > 0
so for any real x, p(x) is never 0
p(x) is irreducible in R (and therefore in Q).
(Note: one could have also used the rational root theorem for irreducibility in Q only.)
/Edit: Cripes! yes, sorry, was too quick. Thumbs down well deserved!