# Abstract algebra: irreducible polynomials

## Homework Statement

Prove that f(x)=x^3-7x+11 is irreducible over Q

## The Attempt at a Solution

I've tried using the eisenstein criterion for the polynomial. It doesn't work as it is written so I created a new polynomial g(x)=f(x+1)=(x+1)^3-7(x+1)+11=x^3+3x^2-4x+5. I did this because g(x) and f(x) are similar and if g(x) is irreducible so is f(x), but the new polynomial I constructed doesn't meet the eisenstein criterion either. Any ideas on where I should turn next?

I've also tried breaking up the polynomial into two smaller ones:
x^3-7x+11=(x+a)(x^2+bx+c)=x^3+(a+b)x^2+(ab+c)x+ac
so from this, I get:
a+b=0; ab+c=-7; ac=11
by rearranging the equations I get b(-b^2+7)=11, but I don't know where I could go from there to show that no solutions for b exist in Q.

Hurkyl
Staff Emeritus
Gold Member
ac=11
Is that a rather severe limitation on a and c?

I think I see what you're saying, a and c can't be integers for their product to be 11, but they could both be rationals. I'm trying to prove that the polynomial is irreducible over Q. I do see that when I put b(-b^2+7)=11 into a calculator, my solution is not a rational number. That justifies that it's not reducible to me, but I'm sure there's a more definitive way of showing that without having to say "my calculator says this".

Dick
Homework Helper
If it's reducible over Q then it must have a rational root, since it's a cubic. You might want to use the 'rational root theorem'.

Hurkyl
Staff Emeritus