Question

Prove that x4 + 3x + 4x² + 8x + 11 is irreducible in Q[x] . Make sure to completely justify all your claims.

Answer 1

We have to show that the polynomial

$\small p(x)=x^4+3x^3+4x^2+8x+11$

is irreducible. Let us first take the polynomial modulo 2, to get

$\small p(x)=x^4+3x^3+4x^2+8x+11 \equiv (x^4+x^3+0x^2+0x+1)\mod 2$

that is, $\small p(x)\equiv (x^4+x^3+1)\mod 2$

Now, this polynomial is of degree 4 over the finite field of two elements, that is, $\small \mathbb F_2=\{0,1\}$

Let us check if any of these two elements are roots of this polynomial, so we have

$\small \text{when }x=1, \Rightarrow (1)^4+(1)^3+1=3\equiv 1\mod 2$

$\small \text{when }x=0, \Rightarrow (0)^4+(0)^3+1=1\equiv 1\mod 2$

Hence, p(x) does not have any roots in F2. So, no linear term in F2 can be a factor of the polynomial. Thus, if it is reducible, it must have quadratic factors in F2, moreover, these quadratic factors cannot have linear factors either.

Now, the 2 degree polynomials is F2[x] are:-

$\small x^2,x^2+1,x^2+x+1$

out of which, we can see that only the last one does not have any roots in F2( 0 is a root of the first one and 1 is a root of the second one )

Thus, the polynomial p(x), if reducible, must be equal to $\small (x^2+x+1)^2$ , however,

$\small (x^2+x+1)^2\equiv (x^4+x^2+1)\mod 2$

Thus, p(x) cannot be factored any further is F2. Thus p(x) is irreducible over F2.

Now, we know that if a polynomial is irreducible over a finite field, it is irreducible over Q too.

Hence, p(x) is irreducible over Q.