Prove that the polynomial $x^3+7x+3$ and $x^3+7x+4$ are irreducible over $\Bbb Q.$

Question

I am a beginner in field extension theory. I need to prove that the polynomials $x^3+7x+3$ and $x^3+7x+4$ are irreducible over $\Bbb Q.$ The only thing that I have learnt in relation to this is Eisenstein's criterion. But I just can't find a suitable prime $p$. Since Eisenstein also works by substitution, I have also tried with replacing $x$ by $x\pm1,x\pm2,x\pm3$. Still that does not work. So I really don't know how to do this then. Can someone help me, please?

Answer 1

If the leading coefficient is $1$ and the degree of the polynomial is less than $4$ (our situation), it is sufficient to prove that the polynomial has no rational root.

The only possible rational roots are the divisors (also the negative ones!) of the constant coefficient in this case.

Answer 2

Start from definition. If $f$ is not irreducible, then $f=gh$ for some non-invertible polynomials $g,h$. Since $\mathbb Q$ is integral domain, $\deg f = \deg g + \deg h$. Since $\deg f = 3$, we conclude that one of $g,h$ has degree $1$ (why??). But, that means that $f(X) = (X-\alpha)(X^2+aX+b)$, which means that $f$ has rational root $\alpha$. Can you show that having rational root is necessary condition as well?

Thus, you only have to check if $f$ has rational roots or not. Use rational root theorem.

Answer 3

Let me do the first polynomial. Call it $f(x)$. Note that $f'(x) = 3x^2+7$. As this derivative is always positive it follows that $f(x)$ is a monotonically increasing function throughout the real line, hence it is a on=one function. Now $f(-1)=-5$ and $f(0)=+3$, and so the unique real root is in between $-1$ and $0$. If we show that this real root is irrational, irreducibility of $f(x)$ will follow.

Write such a rational root as $-a/b$ with $a,b$ positive integers that are coprime with $a<b$.

From $f(-a/b)=0$ it follows that $\displaystyle \frac{a^3}{b^3} +\frac{7a}b=3$. This implies that $a^3=b(3b^2-7ab)$. As rhs is a multiple of $b$ we forces to conclude that $a$ is also a multiple of $b$. That means $a/b$ is an integer. But our root is in between $-1$ and $0$. This contradiction proves that there are no rational roots.