- #1
BSMSMSTMSPHD
- 131
- 0
3. Show that the polynomial [tex]p(x) = x^4 + 5x^2 + 3x + 2[/tex] is irreducible in [tex] \bbmath{Q} [x] [/tex].
This on has me totally stumped. I've seen many other problems on polynomial reducibility, but they are all solvable by Eisenstein, or by checking for irreducibility in a quotient ring by a proper ideal. Neither applies here.
I know that if it does reduce, then there are only two cases to check.
Case 1: [tex]p(x) = a(x)b(x)[/tex] where [tex]deg(a(x)) = 3, deg(b(x)) = 1[/tex].
This case is easy to eliminate, since it implies that [tex]p(x)[/tex] must have a root in [tex]\mathbb{Q}[/tex]. Since the only possible choices are [tex]\pm 1, \pm 2[/tex], it is easy to check that none of these work.
So, Case 1 is impossible.
Therefore, if [tex]p(x)[/tex] is reducible, then it must reduce into two monic polynomials of degree 2.
Case 2: [tex]p(x) = a(x)b(x)[/tex] where [tex]deg(a(x)) = deg(b(x)) = 2[/tex].
Here is where I hit a dead end. I've tried my entire bag of tricks and I have nothing to show for it. Can someone get me started?
This on has me totally stumped. I've seen many other problems on polynomial reducibility, but they are all solvable by Eisenstein, or by checking for irreducibility in a quotient ring by a proper ideal. Neither applies here.
I know that if it does reduce, then there are only two cases to check.
Case 1: [tex]p(x) = a(x)b(x)[/tex] where [tex]deg(a(x)) = 3, deg(b(x)) = 1[/tex].
This case is easy to eliminate, since it implies that [tex]p(x)[/tex] must have a root in [tex]\mathbb{Q}[/tex]. Since the only possible choices are [tex]\pm 1, \pm 2[/tex], it is easy to check that none of these work.
So, Case 1 is impossible.
Therefore, if [tex]p(x)[/tex] is reducible, then it must reduce into two monic polynomials of degree 2.
Case 2: [tex]p(x) = a(x)b(x)[/tex] where [tex]deg(a(x)) = deg(b(x)) = 2[/tex].
Here is where I hit a dead end. I've tried my entire bag of tricks and I have nothing to show for it. Can someone get me started?