Show that polynomial is reducible over $\mathbb{Q}[X]$.

Question

My teacher solved the following problem:

Find out if the polynomial $F(X) = X^3+3X+2$ is irreducible or reducible over $\mathbb{Q}[X]$.

His proof was the following one: The root of the polynomial are of form $\frac{a}{b}$.

$F\left(\frac{a}{b}\right) = \left(\frac{a}{b}\right)^3 + 3\left(\frac{a}{b}\right) + 2 = 0 $ | ( we multiply by $b^3$)

$a^3 +3ab^2 +2b^3 = 0 $

$a^3+b^2(3a+2b) = 0 $

From here he deduced that $b\mid a^3$ and that $a \mid 2$. How did he find this out? Why is this true? From this he further stated that $b=\pm1$ and $a=\pm2$.

Can someone help me out with the deduction that $b \mid a^3$ and $a\mid 2$?

Answer 1

Actually you know from the very beginning that $b \mid a$ since the polynomial is monic and $\mathbb Z$ is integral closed in $\mathbb Q$.

But let's deduce your statements from $a^3 + b^2(3a+2b) = 0$. This equation implies that $a^3$ and $b^2(3a+2b)$ differ only by a sign, so they have the same factors. Since $b$ is a factor of $b^2(3a+2b)$ it has also to be a factor of $a^3$, so $b \mid a^3$.

We can assume $a$ and $b$ to be coprime (by reducing the fraction $a/b$). So $b \mid a^3$ implies $b= \pm 1$. So $a/b \in \mathbb Z$, so without loss of generality $b=1$. Hence $a$ is a root of the polynomial and $(X-a)$ is a factor of the polynomial. The constant terms of $F/(X-a)$ and $X-a$ have to multiply to $2$, so $a \mid 2$.

$F(X) \ge 2$ for $X \ge 0$. So $a \in \{-1,-2\}$. Evaluating gives $a=-2$.

Answer 2

Are you allowed to use the rational root test? If yes, then $a\mid 2$ and $b\mid 1$. A proof can be found here on MSE, e.g., at this question. In any case, it may be useful for the future.

Answer 3

Another way is as follow: $$X^3+3X+2=(X-a)(X^2+bX-c); a,b,c\in\mathbb Z\iff\begin{cases}c+a^2=- 3\\ac=2\end{cases}$$ The only integer solutions for $ac=2$ are not compatible with $c+a^2=-3$.

Thus $F(X)$ is irreducible.