# Show that a polynomial has no irreducible quadratic factors by considering coefficients modulo p

I have a problem that I am working on that I don't understand. I have a degree 4 polynomial that I must show is irreducible in $$\mathbb{Q}[x]$$

$$f(x)=3x^{4}+4x^{3}+6x^{2}+12x+12$$

I have shown using the rational root test that there are no rational roots and hence no linear factors. Then the only factorization of the polynomial would be two quadratic polynomials. The question says to consider the polynomial modulo 3 to show that there are no degree 2 factors, but I don't understand how to do this.

I get:

$$f(x) \equiv x^{3} \mod{3}$$

Could someone explain why this rules out a factor of degree 2?

• I guess you're supposed to write $(ax^2+bx+c)(dx^2+ex+f)\equiv x^3\pmod{3}$ and compare coefficients. Oct 13, 2019 at 15:32

Assume that $$f=gh$$, where $$g, h$$ have degree $$2$$. Since $$f\in\mathbb{Z}[X]$$ is primitive, Gauss lemma shows that we may assume that $$g,h$$ both lie in $$\mathbb{Z}[X]$$ and are primitive. Since the product of leading coefficients of $$g$$ and $$h$$ is $$3$$, one may assume (replacing $$g$$ by $$-g$$ ans $$h$$ by $$-h$$ if nescessary) that $$g$$ is monic and the leading coefficient of $$h$$ is $$3$$.

Reducing mod $$3$$ then prove that $$g\equiv X^2 \mod 3$$ (because $$g$$ is monic) and then $$h\equiv X \mod 3$$. Thus $$g=X^2+3aX+3b, h=3X^2+X+3d$$. The constant term of $$f=gh$$ is then divisible by $$9$$, contradiction.

• Thanks! That was explained really well :)
– user501847
Oct 13, 2019 at 17:51